what is the enthalpy change when a 6.00 g-sample of table sugar (c12h22o11) is oxidized? c12h22o11(s) 12 o2(g) 12 co2(g) 11 h2o(l) h

Answers

Answer 1

The enthalpy change for the combustion of 6.00 g of table sugar is -268,228 kJ/mol.

The enthalpy change for the combustion of one mole of table sugar (C12H22O11) can be calculated using the standard enthalpies of the formation of the reactants and products.

The balanced chemical equation for the combustion of table sugar is:

C12H22O11(s) + 12 O2(g) → 12 CO2(g) + 11 H2O(l)

The standard enthalpies of formation of C12H22O11(s), CO2(g), and H2O(l) are -1274.9 kJ/mol, -393.5 kJ/mol, and -285.8 kJ/mol, respectively. The standard enthalpy of the formation of O2(g) is 0 kJ/mol.

To calculate the enthalpy change for the combustion of 6.00 g of table sugar, we need to convert the mass to moles:

moles of C12H22O11 = 6.00 g / 342.3 g/mol = 0.0175 mol

Using the stoichiometric coefficients in the balanced equation, we can determine that 12 moles of O2 are required to completely react with 1 mole of C12H22O11. Therefore, the number of moles of O2 required to react with 0.0175 mol of C12H22O11 is:

moles of O2 = 12 × 0.0175 mol = 0.21 mol

The enthalpy change for the combustion of 0.21 mol of O2 can be calculated using the standard enthalpies of formation:

ΔH = (12 × -393.5 kJ/mol) + (11 × -285.8 kJ/mol) - (-1274.9 kJ/mol) + (0 kJ/mol)

ΔH = -4694.4 kJ/mol

To calculate the enthalpy change for the combustion of 6.00 g of table sugar, we need to divide by the number of moles of C12H22O11:

ΔH = -4694.4 kJ/mol / 0.0175 mol

ΔH = -268,228 kJ/mol

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Related Questions

aquatic sulfur is primarily: group of answer choices present as the amino acid cysteine present as elemental sulfur present as h2s present as dimethylsulfoniopropionate present as sulfate

Answers

Aquatic sulfur is primarily present as sulfate. Sulfate is the most abundant form of sulfur in seawater, accounting for up to 97 percent of the sulfur present.

Sulfate is important because it is used by microorganisms to perform a variety of biological processes. This includes energy production and the production of organic matter, which is important for supporting marine food webs. Sulfate can also be converted into other forms of sulfur by certain microorganisms.

For example, some bacteria can reduce sulfate to sulfide, which is toxic to most organisms but is used by some bacteria as an energy source. Other microorganisms can oxidize sulfide back to sulfate. Overall, sulfate is an important and abundant form of aquatic sulfur that plays a key role in many marine biological processes.

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Consider the titration of 100.0 mL of 0.100 M H2NNH2 by 0.200 M HNO3. Calculate the pH of the resulting solution after the following volumes of HNO3 have been added.

a. 0.0 mL

pH =

b. 20.0 mL

pH =

c. 25.0 mL

pH =

d. 40.0 mL

pH =

e. 50.0 mL

pH =

f. 100.0 mL

pH =

Answers

the titration of 100.0 mL of 0.100 M H₂NNH₂ by 0.200 M HNO₃

a. pH = 11.94

b. pH = 7.51

c. pH = 7.08

d. pH = 4.57

e. pH = 4.00

f. pH = 1.50

Through the addition of a known quantity of a reactant with a known concentration to the solution until the reaction is complete, the chemical process of titration can be used to determine the concentration of a specific component in a solution. The equivalency point designates the point at which the reaction has ended. When the amount of the substance being studied in the solution is equal to the amount of the added reactant, this is accomplished. To measure the quantity of acids, bases, and other compounds in a sample, analytical chemistry frequently uses titration. An indicator is used in the procedure, and after the reaction is finished, it changes colour, allowing the experimenter to locate the equivalence point. The meticulous measuring of for accurate titration,

A-pH = 14 - pOH = 14 - (-log[OH-]) = 11.94

B-pH = pKa + log([NO3-]/[H2NNH2]) = 8.00 + log(0.0333/0.0800) = 7.51

C-pH = pKa + log([NO3-]/[H2NNH2]) = 8.00 + log(0.0300/0.0625) = 7.08

D-pH = pKa + log([NO3-]/[H2NNH2]) = 8.00 + log(0.0357/0.0500) = 4.57

E-pH = pH of HNO3 solution = 1.18

F-pH = pH of HNO3 solution = 1.00

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how can a lack of natural resources and an abundance of natural resources can each cause problems in society

Answers

A lack of natural resources can cause problems in society, such as economic instability as industries that rely on those resources become unable to operate.

It can also lead to food and water shortages, poverty, and increased inequality.An abundance of natural resources can also cause problems in society. For example, the overuse of resources can lead to environmental degradation, such as deforestation, pollution, and climate change. It can also create economic and political instability, as countries that are heavily reliant on natural resources can struggle to diversify their economies and become overly dependent on a single resource. Additionally, the unequal distribution of these resources can fuel conflict and inequality between countries, regions, and individuals.

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calculate the volume percentage of each major component of the wet bone, based on the weight percentage, i.e. 9, 69, 22 wt% for water, mineral, and organic phase, respectively. assume the densities of the mineral, and organic phase are 3.16 and 1.03 g/cm3, respectively.

Answers

The volume percentage of each major component of wet bone is 9% water, 21.8% mineral, and 21.4% organic phase.

The first step is to calculate the total weight of wet bone, assuming a total weight of 100 grams:

Water: 9 wt% of 100 g = 9 g

Mineral: 69 wt% of 100 g = 69 g

Organic phase: 22 wt% of 100 g = 22 g

The next step is to calculate the volume of each component using their respective densities:

Volume of water = 9 g / 1 g/cm3 = 9 mL

Volume of mineral = 69 g / 3.16 g/cm3 = 21.8 mL

Volume of organic phase = 22 g / 1.03 g/cm3 = 21.4 mL

Finally, we can calculate the volume percentage of each component:

Volume percentage of water = (9 mL / 100 mL) x 100% = 9%

Volume percentage of mineral = (21.8 mL / 100 mL) x 100% = 21.8%

Volume percentage of organic phase = (21.4 mL / 100 mL) x 100% = 21.4%

Therefore, each main component of wet bone has a volume proportion of 9% water, 21.8% mineral, and 21.4% organic phase.

The calculation involves converting the weight percentages to actual weights and then determining the volume of each component based on their respective densities. The volume percentage is then calculated by dividing the volume of each component by the total volume of wet bone (which is assumed to be 100 mL). It is important to use the correct density values for each component to ensure accurate calculations. The volume percentages give us a better understanding of the relative contributions of each component to the overall structure of wet bone.

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at 15 degress celcius, the water ionization constant, k2 is 4.5 * 10^-15. what is th h3o concentration in neutral water at this temperature

Answers

The H3O+ concentration of neutral water at 15 degrees Celsius is 10^-7 M.

As per the given information, the water ionization constant k2 is 4.5 × 10^-15 at 15 degrees Celsius. The H3O+ concentration of neutral water at this temperature can be determined using the formula:

Kw = [H3O+][OH-]Here, Kw represents the ionic product of water which is equal to 1.0 × 10^-14 at 15 degrees Celsius (which can be looked up in a table).

Since we know that pure or neutral water has an equal concentration of H3O+ and OH- ions, we can assume that the concentration of OH- ions is also 1.0 × 10^-7 M.

Therefore, we can substitute the values in the equation to obtain:

[H3O+][OH-] = 1.0 × 10^-14[H3O+] [1.0 × 10^-7]

= 1.0 × 10^-14[H3O+]

= 1.0 × 10^-7 ÷ 1.0 × 10^-14[H3O+]

= 10^7The H3O+ concentration of neutral water at 15 degrees Celsius is 10^-7 M (or 0.0000001 M).

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when two half-cells are connected to form a galvanic cell, a potential of 0.98 v is measured. in order to run this as an electrolytic cell, what potential must be applied?

Answers

to run the same reaction as an electrolytic cell, an external voltage of at least -0.98 V must be applied.

In a galvanic cell, electrical energy is generated spontaneously from a chemical reaction, while in an electrolytic cell, electrical energy is used to drive a nonspontaneous chemical reaction.

The potential difference between the two half-cells in a galvanic cell is known as the cell potential or electromotive force (EMF). In this case, the measured potential is 0.98 V.

To run the same reaction as an electrolytic cell, an external voltage must be applied that is greater than the cell potential. This external voltage is known as the overpotential and is necessary to drive the reaction in the non-spontaneous direction.

The overpotential required depends on the specific reaction and the conditions of the electrolytic cell, such as the concentration of the electrolyte, temperature, and pressure. In general, the overpotential required for an electrolytic cell is equal in magnitude to the cell potential but with the opposite sign.

Therefore, to run the same reaction as an electrolytic cell, an external voltage of at least -0.98 V must be applied.
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a construction company used asbestos on many job sites over the years. asbestos is a(n) group of answer choices widely used dye. pesticide. radioactive material. environmental carcinogen.

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Asbestos is not a widely used dye, pesticide, or radioactive material. As mentioned earlier, asbestos is a group of naturally occurring fibrous minerals that were widely used in construction and insulation materials due to their heat resistance and durability.

However, prolonged exposure to asbestos fibres can lead to serious health problems, including lung cancer, mesothelioma, and asbestosis. These health risks are due to the fact that asbestos fibres are small and can be inhaled, causing damage to the lungs and respiratory system. Therefore, asbestos is classified as an environmental carcinogen, which means it is a substance that can cause cancer or promote the growth of cancer cells in the environment.

The construction company's use of asbestos on job sites could potentially have exposed workers and others to these health risks, which is why the use of asbestos has been heavily regulated and phased out in many countries around the world.

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a vinegar sample requires 41.30 ml of a 0.1042 m naoh solution to reach the phenolphthalein endpoint. calculate the moles of acetic acid present in the vinegar sample.

Answers

Given, Volume of vinegar sample = 41.30 ml Concentration of NaOH = 0.1042 m Moles of acetic acid present in the vinegar sample can be calculated as shown below : NaOH + CH3COOH → CH3COONa + H2OThus, 1 mole of NaOH reacts with 1 mole of CH3COOH.The volume of NaOH required to neutralize CH3COOH present in the vinegar sample is 41.30 ml.

Hence, the number of moles of NaOH used can be calculated as follows: Number of moles of NaOH = (Concentration of NaOH) x (Volume of NaOH used in L)Number of moles of NaOH = (0.1042 x 10^-3) x (41.30 x 10^-3)Number of moles of NaOH = 4.30 x 10^-3 moles. Thus, the number of moles of CH3COOH present in the vinegar sample is 4.30 x 10^-3 moles.

Therefore the answer is: the number of moles of CH3COOH present in the vinegar sample is 4.30 x 10^-3 moles.

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if you have 100 g of naoh and 100 g of al to perform the reaction, how many grams of h2 will you produce?

Answers

2.5 grams of [tex]H_{2}[/tex] will produce to perform the reaction.

[tex]2Al+2NaOH+2K_{2} O[/tex]   ⇒  [tex]2Na_{3} AlO_{3} +3H_{2}[/tex]

The reactants have the following molar masses: M(NaOH) = 40 g/mol, and M(Al) = 27 g/mo.

Given,

100 g of NaOH.

100 g of Al

The reactants' total moles are:

() =()/() = 100/27 = 3.7

() = ()/() = 100/40 = 2.5 l

Then we calculate a ratio:

6 moles of NaOH and 2 moles of Al react,

3.7 moles of Al should react with x moles of NaOH, where x = 3.7*  6 /2 = 11.1 moles of NaOH.

The limiting reactant is sodium hydroxide as there are only 2.5 moles of it. To determine the mass of H2 that could be created by the chemical reaction, we must use a different percentage.

3 moles of H2 are produced from 6 moles of NaOH.

NaOH 2.5 moles - x moles H2

=2.5* 3/6 = 1.25 2

mass of [tex]H_{2}[/tex] = n([tex]H_{2}[/tex]) * M([tex]H_{2}[/tex]) = 1.25* 2 = 2.5 gm

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when the student read the eudiometer tube to record the volume of h2 gas, he read the edge of the meniscus instead of the center. how will this affect the measured volume of the gas?

Answers

Reading the edge of the meniscus instead of the center of the meniscus will result in an inaccurate measurement of the volume of the gas.

The meniscus should be read at the middle of the curve, not the edge, when using the eudiometer tube.

The volume of the gas will be more than the real volume if the meniscus' edge is read rather than the centre since the edge is higher than the centre.

The eye should be level with the meniscus's centre when reading the tube for the most precise measurement. Inaccurate calculations and faulty findings can occur from measuring the gas's volume incorrectly.

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what is the concentration of h in a 0.025 m hcl solution? group of answer choices 0.010 m 0.013 m 0.025 m 0 0.050 m

Answers

Answer:

0.025

Explanation:

The concentration is 0.025 since that is the molarity of the solution. Molarity is mol/L . Molarity is the concentration of the HCl

Answer is 0.025

The concentration of H+ in a 0.025 M HCl solution is 0.025 M, since HCl is a strong acid and dissociates completely in water to form H+ and Cl-.

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calculate the ph of a buffer solution obtained by dissolving 30.0 g of kh2po4(s) and 49.0 g of na2hpo4(s) in water and then diluting to 1.00 l.

Answers

The pH of the buffer solution is approximately 7.65.

To calculate the pH of a buffer solution, first determine the concentration of [tex]KH_{2} PO_{4}[/tex] and [tex]Na_{2} HPO_{4}[/tex].

Calculate moles of [tex]KH_{2} PO_{4}[/tex] and [tex]Na_{2} HPO_{4}[/tex]:


- [tex]KH_{2} PO_{4}[/tex]: (30.0 g) / (136.09 g/mol) = 0.2203 mol


- [tex]Na_{2} HPO_{4}[/tex]: (49.0 g) / (141.96 g/mol) = 0.3451 mol

Calculate concentrations in the 1.00 L solution:


- [[tex]KH_{2} PO_{4}[/tex]] = 0.2203 mol / 1.00 L = 0.2203 M


- [[tex]Na_{2} HPO_{4}[/tex]] = 0.3451 mol / 1.00 L = 0.3451 M

Use the Henderson-Hasselbalch equation:

pH = pKa + log ([A-]/[HA])


- For [tex]H_{2} PO_{4}[/tex]-/HPO4²-, the pKa = 7.21.


- pH = 7.21 + log (0.3451/0.2203)

Calculate the pH:


- pH ≈ 7.21 + 0.438 = 7.65

The pH of the buffer solution is approximately 7.65.

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A container holds three gases at a total pressure of 800 kPa. If the partial pressure of the first gas is 100 kPa and the partial pressure of the second gas is 300 kPa, what is the partial pressure of the third gas?

Answers

We can make use of the fact that the total pressure of the mixture is the same as the sum of the partial pressure of all the gases in the container. As a result, 400 kPa is the third gas' partial pressure.

What exactly is a partial pressure law?

The overall pressure exerted by a mixture of gases is equal to the sum of the partial pressures of each of the constituent gases, according to Dalton's law of partial pressures. The partial pressure is the pressure that each gas would have if it occupied the same volume of the mixture at the same temperature on its own.

Mathematically, this can be stated as:

P_total = P_1 + P_2 + P_3

When we plug these figures into the formula, we get:

800 kPa = 100 kPa + 300 kPa + P_3

When we simplify this equation, we obtain:

800 kPa = 400 kPa + P_3

By taking away 400 kPa from both sides, we arrive at:

P_3 = 400 kPa

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Find the mole value of 68 liters of O2

Answers

68 liters of O₂ at STP is equivalent to 2.693 moles of O₂.

To find the mole value of 68 liters of O₂, we need to use the ideal gas law:

PV = nRT

where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature of the gas.

Assuming that the O₂ is at standard temperature and pressure (STP), which is 0°C and 1 atm, respectively, we can use the following values for the variables in the ideal gas law:

P = 1 atm

V = 68 L

n = ?

R = 0.08206 L·atm/(mol·K)

T = 273 K

Substituting these values into the ideal gas law and solving for n, we get:

n = PV/RT

n = (1 atm) * (68 L) / (0.08206 L·atm/(mol·K) * 273 K)

n = 2.693 moles

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What are the similarities between structural models and Lewis dot structures?

Answers

Answer:

A Lewis dot structure is like a simplified Bohr-Rutherford model. The Lewis Dot diagram contains the element symbol with dots representing electrons. The only electrons shown are those on the outer energy level or valence electrons.

Finish the sentence. Cold air molecules heat up when they touch the warm ground through the process of ______________.

Density


Radiation


Conduction


Evaporation

Answers

Answer:

Conduction

Explanation:

Conduction is the transfer of heat energy from one substance to another or within a substance.

Cold air molecules heat up when they touch the warm ground through the process of  conduction.

What is conduction ?

Conduction is the transfer of heat or energy between two objects that are in contact with each other, or between different parts of the same object, due to a temperature difference.

In conduction, heat energy is transferred through a material or substance from higher-temperature regions to lower-temperature regions.

Conduction occurs because the molecules in a substance are in constant motion and collisions between them can transfer energy. When two objects are in contact, the faster-moving molecules in the warmer object collide with the slower-moving molecules in the cooler object, transferring energy from the warmer object to the cooler one.

This process continues until the two objects reach thermal equilibrium, meaning they have the same temperature.

Materials that are good conductors of heat, such as metals, allow energy to be transferred quickly through them. On the other hand, materials that are poor conductors of heat, such as plastics or insulators, prevent or slow down the transfer of energy.

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How many moles of NaOH are contained in a 65.0 mL of a 2.20 M solution of NaOH?

Answers

Answer:.143 moles

Explanation:

Explanation:

We're going to use this equation

n

=

M

V

M meaning our molarity, n meaning our moles involved, and v meaning our volume in Liters.

n

=

2.20

.065

n=.143 moles.

Answer:

there are 0.143 moles of NaOH in 65.0 mL of a 2.20 M solution of NaOH.

Explanation:

To calculate the number of moles of NaOH in a given solution, we need to use the formula:

moles = concentration (in M) x volume (in L)

We are given the concentration of NaOH solution as 2.20 M and the volume as 65.0 mL. However, we need to convert the volume to liters before we can use the formula.

65.0 mL = 65.0/1000 L = 0.065 L

Now we can plug in the values in the formula:

moles = 2.20 M x 0.065 L

moles = 0.143 moles

Therefore, there are 0.143 moles of NaOH in 65.0 mL of a 2.20 M solution of NaOH.

a mineral deposit along a strip of length 1 cm has density g/cm for . calculate the total mass of the deposit. your answer must include units.

Answers

The total mass of the deposit along a strip of length 1 cm that has density g/cm is simply g.

In order to calculate the total mass of a mineral deposit along a strip of length 1 cm that has density g/cm, we can use the formula: mass = density x volume The volume of the deposit is equal to the product of its length, width, and height. Since the deposit is along a strip of length 1 cm, we can assume that its width and height are negligible. Therefore, the volume of the deposit is approximately 1 cm³.

Substituting the density of the deposit in place of g/cm³, we can rewrite the formula as: mass = (density in g/cm³) x (volume in cm³)mass = g x 1 cm³mass = g To express the mass of the deposit in units of grams (g), we must know the value of the density in g/cm³. Once we know this value, we can multiply it by 1 cm³ to obtain the mass in grams.

For example, if the density of the deposit is 2 g/cm³, then the mass of the deposit is: mass = 2 g/cm³ x 1 cm³mass = 2 g Therefore, the total mass of the mineral deposit along a strip of length 1 cm that has density g/cm is simply g.

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which of the following statement about atoms is true? group of answer choices normally, an atom has an equal number of protons and electrons. all atoms in the universe have the same number of electrons. all atoms in the universe have the same number of protons. any molecule that has mass is an atom. atoms are the simplest part of a proton or a neutron

Answers

Atoms is defined as a particle of matter that consists of a central nucleus that is surrounded by one or more negatively charged electrons which has equal number of protons and electrons. So option (a) id correct.

Atoms are defined as the smallest part of a substance which cannot be broken down chemically. Each of the atom has a nucleus that is present in the center which is made up of protons which are called as positive particles and neutrons that are the particles with no charge. Electrons  are the negative particles which moves around the nucleus. An atom has an equal number of electrons and protons in its cell. It is evident that  atom is electrically neutral because of the number of protons is always equal to a number of electrons. Atoms of the same element  must have the same atomic number so that the number of protons and electrons are being same.  

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which of the following statement about atoms is true?

group of answer choices normally,

A. an atom has an equal number of protons and electrons.

B. all atoms in the universe have the same number of electrons.

C. all atoms in the universe have the same number of protons.

D. any molecule that has mass is an atom.

E. atoms are the simplest part of a proton or a neutron

Help what's the answer??

Answers

Answer:

The most significant quantity of CO2 that can be produced:

The reaction's limiting reactant determines the maximum amount of carbon dioxide that can be generated. The limiting reactant is the reactant that is totally consumed in the reaction and dictates how much product may be generated. To identify the limiting reactant, compare the available quantities of glucose and oxygen and calculate the carbon dioxide that can be generated from each.

Glucose (C6H12O6) has a molar mass of 180.18 g/mol.

Oxygen (O2) has a molar mass of 32.00 g/mol.

Explanation:

We can compute the number of moles of each reactant using these values:

Moles of glucose = 9.91 g / 180.18 g/mol = 0.055 mol

Amount of oxygen moles = 15.2 g/32.00 g/mol = 0.475 mol

The reaction's balanced chemical equation is:

C6H12O6 + 6O2 + 6CO2 = C6H12O6 + 6H2O

One mole of glucose combines with six moles of oxygen to generate six moles of carbon dioxide, according to the balanced equation. As a result, the greatest quantity of carbon dioxide that may be generated from the given reactant quantities is:

The highest quantity of carbon dioxide that may be generated if glucose is the limiting reactant is: 0.055 mol glucose 6 mol CO2/mol glucose 44.01 g/mol CO2 = 16.93 g CO2

The highest quantity of carbon dioxide that may be generated if oxygen is the limiting reactant is: 0.475 mol oxygen 1 mol CO2/mol oxygen 44.01 g/mol CO2 = 20.89 g CO2

As a result, the maximum quantity of CO2 that may be produced is 16.93 g CO2 (when glucose is the limiting reactant).

The following is the formula for the limiting reactant:

To get the limiting reactant formula, we must compute the theoretical production of carbon dioxide for each reactant and compare the results. The limiting reactant is the one that produces the least quantity of carbon dioxide.

Based on glucose as the limiting reactant, the potential production of carbon dioxide is: 0.055 mol glucose 6 mol CO2/mol glucose 44.01 g/mol CO2 = 16.93 g CO2

Based on oxygen as the limiting reactant, the potential production of carbon dioxide is: 0.475 mol oxygen 1 mol CO2/mol oxygen 44.01 g/mol CO2 = 20.89 g CO2

As a result, glucose is the limiting reactant, with the formula C6H12O6.

The amount of extra reactant that remains:

Excess reactants are those that are not totally consumed in the reaction. To compute the quantity of surplus reactant left, we must first estimate how much oxygen interacts with glucose. According to the balanced equation, one mole of glucose combines with six moles of oxygen. As a result, the number of moles of oxygen reacting with glucose is:

6 mol O2/mol glucose = 0.055 mol glucose = 0.33 mol O2

The quantity of oxygen remaining after the reaction is as follows:

15.2 g of O2 - 0.33 mol of O2 32.00 g/mol = 4.76 g of O2.

As a result, when the reaction is complete, 4.76 g of oxygen remains as the surplus reactant.

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consider the following oxides: so2 , p2o5 , mgo , cl2o , n2o5 . how many are expected to form acidic solutions in water?

Answers

Metal oxides form basic solutions while nonmetal oxides form acid solutions. SO₂, P₂O₅, Cl₂O, and N₂O₅ oxides are predicted to generate acidic solutions.

The periodic table's left side is where metals are located. Oxides are created when metals and oxygen react. Metal oxides disintegrate in water to create basic solutions. The oxides that should result in the formation of a basic solution are MgO,

Oxygen and nonmetals can also react to generate oxides. Nonmetal oxides dissolve in water to produce acids. Hence, SO₂, P₂O₅, Cl₂O, and N₂O₅ oxides are predicted to generate acidic solutions.

Metal oxides have a basic character, whereas non-metal oxides have an acidic nature. Metal oxides are basic because when they dissolve in water, they produce salt and water. Metal oxides include OH⁻. Ions are thus fundamental. H⁺ ions are found in non-metal oxides like sulphur dioxide, which dissolve in water to create acidic solutions.

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An unknown gas diffuses 0.25 times as fast as He, What is the molecular mass of the
unknown gas?

Answers

The molecular mass of the unknown gas = 64g/mol

The rate of diffusion of a gas is inversely proportional to the square root of its molecular mass. Let the molecular mass of the unknown gas be M. Then, we have:

Rate of diffusion of unknown gas / Rate of diffusion of He = sqrt(MHe / M)

Since the rate of diffusion of the unknown gas is 0.25 times that of He, we can write:

0.25 = sqrt(MHe / M)

Squaring both sides, we get:

0.0625 = MHe / M

M = MHe / 0.0625

The molecular mass of He is approximately 4 g/mol. Substituting this value, we get:

M = 4 / 0.0625

M = 64 g/mol

Therefore, the molecular mass of the unknown gas is approximately 64 g/mol.

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a 50.0- ml volume of 0.15 m hbr is titrated with 0.25 m koh . calculate the ph after the addition of 17.0 ml of koh .

Answers

The pH after the addition of 17.0 mL of KOH is 0.602.

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Here is the solution to the given problem:Given:Volume of HBr = 50.0 mL

Concentration of HBr = 0.15M

Concentration of KOH = 0.25 M

Volume of KOH added = 17.0 mL

We need to calculate the pH after the addition of KOH.The balanced chemical equation for the reaction between HBr and KOH is:

HBr + KOH → KBr + H2O

Initial moles of HBr = Molarity × Volume

= 0.15 mol/L × (50.0 mL/1000)

= 0.0075

molInitial moles of KOH = Molarity × Volume

= 0.25 mol/L × (17.0 mL/1000) = 0.00425

 sinceNonsense the reaction is a neutralization reaction between a strong acid and a strong base, we can assume that the volume after the reaction would be 50 + 17 = 67 mL or 0.067 L.

The moles of HBr left after reaction = initial moles of HBr - moles of KOH reacted

= 0.0075 - 0.00425= 0.00325 mol

Concentration of HBr after reaction

= Moles of HBr/Volume of solution = 0.00325 mol/0.067

L = 0.0485 MConcentration of H3O+ = Concentration of OH-

= Molarity of KOH added

= 0.25 MSo,

the pH can be calculated using the formula:

pH = -log[H3O+]= -log[0.25]

= 0.602

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ammonia chemically reacts with oxygen gas to produce nitric oxide and water . what mass of nitric oxide is produced by the reaction of of oxygen gas?

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The mass of oxygen gas is 5.77 g. Hence, the mass of nitric oxide produced by the reaction of 5.77 g of oxygen gas is 4.32 g.

They want to know the mass of nitric oxide that is produced by this reaction. The balanced chemical equation for this reaction is:

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

In this equation, 4 moles of NH₃ react with 5 moles of O₂ to produce 4 moles of NO and 6 moles of H₂O. To find the mass of NO produced, we need to use the molar mass of NO. Molar mass of NO = 30.01 g/molTo find the mass of NO produced, we need to use stoichiometry.

We know that 4 moles of NH₃ reacts with 5 moles of O₂ to produce 4 moles of NO. Therefore, if we know the number of moles of O₂ that reacted, we can use stoichiometry to find the number of moles of NO produced.We can use the ideal gas law to find the number of moles of O₂ that reacted.

n = PV/RT  We can assume that the reaction took place at standard temperature and pressure (STP), which means:P = 1 atmV = 22.4 L (molar volume of an ideal gas at STP)T = 273.15 K (0 °C)R = 0.08206 L atm/(mol K)Using these values, we can find the number of moles of O₂ that reacted:n = PV/RT = (1 atm)(22.4 L)/(0.08206 L atm/(mol K) * 273.15 K) ≈ 1 mol

Therefore, 1 mole of O₂ reacted in the reaction. Using stoichiometry, we can find the number of moles of NO produced.4 moles NH₃ : 4 moles NO1 mole O₂ : 4/5 moles NO (from the balanced equation)1 mole O₂ was consumed, so the number of moles of NO produced is:1 mole O₂ * (4/5 moles NO/1 mole O₂) = 0.8 moles NO

Finally, we can find the mass of NO produced using the molar mass of NO:mass NO = number of moles * molar mass mass NO = 0.8 mol * 30.01 g/mol ≈ 24.0 g Therefore, approximately 24.0 grams of NO are produced by the reaction between ammonia and oxygen gas.

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agi , baf2 , and agbr are all sparingly soluble salts. which of these salts will be more soluble in an acidic solution than in water?

Answers

sparingly soluble salts derived from weak acids tend to be more soluble in an acidic solution.

This is because the added H+ ions (from a strong acid such as HCl) react with the anions of the salts to form weak acids, which decrease the ion product and shift the equilibrium to favor more dissolution. Therefore, out of the three salts you mentioned, BaF_2 will be more soluble in an acidic solution than in water. This is because F- is the conjugate base of HF, a weak acid, while Ag+ and Br- are the conjugate acids of strong bases (AgOH and KOH).

The solubility product constant (Ksp) of BaF_2 is a measure of how much BaF_2 can dissolve in water at a given temperature. It is defined as the product of the concentrations of Ba^2+ and F^- ions in equilibrium with solid BaF_2. According to Bartleby1, the solubility product constant for BaF_2 is 1.0 x 10^-6 at 25°C. This means that only a very small amount of BaF_2 can dissolve in water at this temperature.

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Consider the titration of 100.0 mL of 0.100 M H2NNH2 (Kb = 0.000003) by 0.200 M HNO3. Calculate the pH of the resulting solution after the following volumes of HNO3 have been added.

a. 0.0 mL

pH =

b. 20.0 mL

pH =

c. 25.0 mL

pH =

d. 40.0 mL

pH =

e. 50.0 mL

pH =

f. 100.0 mL

pH =

Answers

Answer:

A = 11.74
B = 9.18
C = 10.4
D = 8.10
E = 7.21
F = 1.70

For more context, you may check the explanation :)

Explanation:

This is a basic titration problem where an amine, H2NNH2, is being titrated with an acid, HNO3. The reaction between the two is:

H2NNH2 + HNO3 → H2NNH3+NO3-

The Kb value for H2NNH2 is given as 0.000003, which allows us to calculate the Kb expression:

Kb = [H2NNH3+][OH-] / [H2NNH2]

At the start of the titration, before any HNO3 is added, we have only H2NNH2 in solution and no H2NNH3+ or OH-. Therefore, at the start of the titration:

Kb = [H2NNH3+][OH-] / [H2NNH2]

0.000003 = (x)(x) / (0.100)

x = 0.0055 M

So at the start of the titration, [H2NNH2] = 0.100 M and [OH-] = 0.0055 M. To find the pH, we can use the fact that:

pH + pOH = 14

a. Before any HNO3 is added, the [OH-] is 0.0055 M. Therefore:

pOH = -log(0.0055) = 2.26

pH = 14 - 2.26 = 11.74

The pH of the solution is 11.74.

b. At 20.0 mL of HNO3 added, we can calculate the moles of HNO3 added:

moles of HNO3 = (0.200 M)(0.020 L) = 0.004 mol

This amount of HNO3 reacts completely with the same amount of H2NNH2, so the moles of H2NNH2 remaining is:

moles of H2NNH2 = 0.100 mol - 0.004 mol = 0.096 mol

The total volume of the solution is now 0.100 L + 0.020 L = 0.120 L. Therefore, the concentration of H2NNH2 is:

[H2NNH2] = 0.096 mol / 0.120 L = 0.800 M

Using the Kb expression, we can find the [OH-]:

Kb = [H2NNH3+][OH-] / [H2NNH2]

0.000003 = (x)(0.004) / (0.800)

x = 0.000015 M

Therefore, the pOH is:

pOH = -log(0.000015) = 4.82

And the pH is:

pH = 14 - 4.82 = 9.18

The pH of the solution is 9.18.

c. At 25.0 mL of HNO3 added, we can use the same approach as above to find that the concentration of H2NNH2 is 0.625 M, the [OH-] is 0.00004 M, and the pH is 10.4.

d. At 40.0 mL of HNO3 added, we can use the same approach as above to find that the concentration of H2NNH2 is 0.200 M, the [OH-] is 0.00080 M, and the pH is 8.10.

e. At 50.0 mL of HNO3 added, we can use the same approach as above to find that the concentration of H2NNH2 is 0.100 M, the [OH-] is 0.00155 M, and the pH is 7.21.

f. At 100.0 mL of HNO3 added, we can use the same approach as above to find that the concentration of H2NNH2 is 0.0 M, the [OH-] is 0.02000 M, and the pH is 1.70.

8. The reaction of hydrogen with chlorine to form gaseous hydrogen chloride is strongly exothermic. From this it can be deduced that A the temperature falls during the reaction. heat is taken in during the reaction. B C more molecules are formed than are used up in the reaction. D hydrogen chloride gas is not easily decomposed by heat.​

Answers

Answer: The correct option is A) the temperature decreases during the reaction.

Explanation: When hydrogen reacts with chlorine to form hydrochloric acid, the reaction is highly exothermic. This means that the reaction releases energy in the form of heat. The energy released is greater than the energy required to break the bonds in the reactants. Thus, the temperature of the system decreases during the reaction.

Option B is wrong because when heat is absorbed in a reaction, it is an endothermic reaction where energy is absorbed as heat.

Option C is wrong because the number of molecules produced is equal to the number of molecules consumed in the chemical reaction, according to the law of conservation of mass.

Variant D is unrelated to the exothermic nature of the reaction between hydrogen and chlorine leading to the formation of hydrochloric acid. Describes the stability of hydrochloric acid gas unrelated to the statement.

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when isopropylbenzene (cumene) is treated with nbs and irradiated with uv light, only one product is obtained. propose a mechanism and explain why only one product is formed.

Answers

When isopropylbenzene (cumene) is treated with NBS (N-bromosuccinimide) and irradiated with UV light, a radical bromination reaction occurs, resulting in the formation of only one product: p-bromoisopropylbenzene.

The mechanism for this reaction involves three steps:

1. Initiation: UV light irradiation causes homolytic cleavage of the N-Br bond in NBS, generating a bromine radical (Br•).

2. Propagation: The bromine radical abstracts a hydrogen atom from the isopropyl group's benzylic position, forming a benzylic radical. The benzylic radical then reacts with another NBS molecule, forming a new N-Br bond and regenerating the bromine radical.

3. Termination: Radical species combine to form a stable compound, ending the reaction.

Only one product is formed because the benzylic radical formed is resonance-stabilized and can only react with the bromine radical at the para position, relative to the isopropyl group.

This leads to the formation of p-bromoisopropylbenzene, which is the only product observed in this reaction.

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of the following, the entropy of gaseous is the largest at 25ec and 1 atm. a) c2h2 b) h2 c) c2h6 d) ch4 e) c2h2

Answers

The entropy of gaseous H₂ (hydrogen gas) is the largest at 25°C and 1 atm. Option B is correct.

Entropy is a measure of the disorder or randomness of a system. In general, the entropy of a substance increases as it goes from a solid to a liquid to a gas. At the same temperature and pressure, the entropy of a substance is proportional to the number of particles in the system.

H₂ has the smallest molecular weight, which means that it has the greatest number of particles per unit volume compared to the other molecules. Therefore, it has the largest entropy at a given temperature and pressure.

Furthermore, H₂ is a diatomic gas, meaning that it has two atoms per molecule. Diatomic gases have more degrees of freedom than monoatomic gases, which also increases their entropy. C₂ H₂ (acetylene) is also a diatomic gas, but it has a larger molecular weight than H₂, so it has a lower entropy. In summary, the entropy of gaseous H₂ is the largest at 25°C and 1 atm due to its small molecular weight and diatomic nature. Option B is correct.

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A gas absorbs 5.1 kJ of heat and does 3.6 kJ of work. Calculate ΔE.

Answers

A gas absorbs 5.1 kJ of heat and does 3.6 kJ of work then the change in internal energy of the gas is 8.7 kJ.

The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. Mathematically, this can be expressed as:

ΔE = Q - W

where ΔE is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.

In this problem, Q = 5.1 kJ (heat absorbed by the gas) and W = -3.6 kJ (work done by the gas, with the negative sign indicating work done on the surroundings). Plugging these values into the formula, we get:

ΔE = Q - W

ΔE = 5.1 kJ - (-3.6 kJ)

ΔE = 5.1 kJ + 3.6 kJ

ΔE = 8.7 kJ

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