The half‑equivalence point of a titration occurs half way to the equivalence point, where half of the analyte has reacted to form its conjugate, and the other half still remains unreacted. If 0.480 moles of a monoprotic weak acid (Ka=3.0×10−5) is titrated with NaOH, what is the pH of the solution at the half‑equivalence point?

Answer:

Oxidizing agent - CrO4^2-

Reducing agent- N2O

Explanation:

Let us look at the equation closely;

CrO4^2- (aq) + 3N2O(g) ------------> Cr^3+ (aq) + 3NO(g) [acidic]

The reduction half equation is;

CrO4^2- (aq) + 3e -------->Cr^3+ (aq)

Oxidation half equation is;

3N2O(g) ------>3 NO(g) +3 e

Note that the oxidizing agent participates in the reduction half equation while the reducing agent participates in the oxidation half equation as seen above.

Answer:

8.96g\ cm3

Explanation:

D = ( 89.6g \ 10cm3)

( 89.6\ 10) ( g\ cm3) = 8.96g\cm3

Answer:

-179.06 kJ

Explanation:

Let's consider the following balanced reaction.

HCl(g) + NaOH(s) ⟶ NaCl(s) + H₂O(l)

We can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.

ΔH°r = 1 mol × ΔH°f(NaCl(s)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(NaOH(s))

ΔH°r = 1 mol × (-411.15 kJ/mol) + 1 mol × (-285.83 kJ/mol) - 1 mol × (-92.31 kJ/mol) - 1 mol × (-425.61 kJ/mol)

ΔH°r = -179.06 kJ

Explanation:

Combustion of butane gives following reaction

C4H10 + 13/2 O2 → 5H20 + 4CO2

Therefore, One Kg of Butane consists of 1000/58 moles . That is 17.24 moles.

17.24 moles of butane reacts with (6.5×17.24) moles of O2 = 112.06 moles = 3.586 Kg O2.

Weight % of O2 in Air = 20 %

Mass of theoretical air used is 3.586×5 = 17.93 Kg = 71.72 %

The dew point temperature of H2O is :

D = (237.3×B)/(1-B)

B = ln(E/6.108)/17.27

E is vapor pressure at given temp.

At T = 30° C , E = 31.8 mm of Hg = 41.84 Milli Bars

B = 0.1114

D = 29.74°C

Answer:

[tex]m_{Si}=37.2gSi[/tex]

Explanation:

Hello,

In this case, for the undergoing balanced chemical reaction:

[tex]SiCl_4 + 2Mg \rightarrow Si + 2MgCl_2[/tex]

We must first identify the limiting reactant given the 225 g of SiCl4 and 101 g of Mg. Thus, we compute the available moles of SiCl4:

[tex]n_{SiCl_4}=225gSiCl_4*\frac{1molSiCl_4}{169.8963gSiCl_4}=1.324molSiCl_4[/tex]

Next, by using the 1:2 mole ratio between SiCl4 and Mg, we compute the moles of SiCl4 consumed by 101 g of Mg:

[tex]n_{SiCl_4}^{consumed}=101gMg*\frac{1molMg}{24.3050gMg} *\frac{1molSiCl_4}{2molMg} =2.08molSiCl_4[/tex]

Thus, since less moles of SiCl4 are available, we can infer it is the limiting reactant whereas the Mg is in excess. In such a way, the produced grams of Si are computed considering the 1:1 molar ratio between SiCl4 and Si:

[tex]m_{Si}=1.324molSiCl_4*\frac{1molSi}{1molSiCl_4} *\frac{28.0855gSi}{1molSi} \\\\m_{Si}=37.2gSi[/tex]

Best regards.

Answer:

The correct answer is 1.60.

Explanation:

Based on the given question, 25 ml of 0.10 M HCl is titrated with 0.10 M NaOH. Now moles of HCl can be determined by using the formula,  

Moles = volume * concentration of HCl

= 25/1000*0.10 = 0.0025 moles

Similarly the moles of NaOH added will be determined by using the formula,  

Moles of NaOH added = volume * concentration of NaOH

= 15/1000 * 0.10 = 0.0015 moles

The reaction taking place in the given case is,  

HCl + NaOH = NaCl + H2O

Now the moles of excess H+ = moles of excess HCl

= 0.0025 - 0.0015 = 0.001 moles

Based on the given question, the sum of the volume of the solution is 25+15 = 40 ml or 0.040 L

[H+] = moles of H+/total volume

= 0.001 / 0.040 = 0.025 M

pH = -log[H+]

= -log[0.025]  

= 1.60

The pH after 15 ml of NaOH has been volume is 1.60.

It is based on the given question that is, 25 ml of 0.10 M HCl is titrated with 0.10 M NaOH. Now moles of HCl can be specified by using the formula,

Moles is = volume * concentration of HCl

Then is = 25/1000*0.10 = 0.0025 moles

Besides, the moles of NaOH added will be determined by using the formula,

When the Moles of NaOH added is = volume * concentration of NaOH

= 15/1000 * 0.10 = 0.0015 moles

When The reaction taking place in the given case is,

HCl + NaOH = NaCl + H2O

Now the moles of excess H+ = moles of excess HCl

= 0.0025 - 0.0015 = 0.001 moles

Then It Based on the given question, the sum of the volume of the solution is 25+15 = 40 ml or 0.040 L

[H+] = moles of H+/total volume

After that = 0.001 / 0.040 = 0.025 M

pH = -log[H+]

Then = -log[0.025]

Therefore, = 1.60

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Answer:

Q = 246 kJ

Explanation:

It is given that,

Mass of water, m = 200 g

Let initial temperature, [tex]T_i=5^{\circ}[/tex]

Final temperature of water, [tex]T_f=300^{\circ} C[/tex]

We know that the specific heat capacity of water, [tex]c=4.18\ J/g-^{\circ} C[/tex]

So, the heat energy needed to raise the temperature is given by :

[tex]Q=mc\Delta T\\\\Q=200\times 4.18\times (300-5)\\\\Q=246620\ J[/tex]

or

Q = 246 kJ

So, the heat energy of 246 kJ is needed.

Answer:

28.0mL of the 0.0500M NaOH solution

Explanation:

0.126g of lactic acid diluted to 250mL. Titrated with 0.0500M NaOH solution.

The reaction of lactic acid, H₃C-CH(OH)-COOH (Molar mass: 90.08g/mol) with NaOH is:

H₃C-CH(OH)-COOH + NaOH → H₃C-CH(OH)-COO⁻ + Na⁺ + H₂O

Where 1 mole of the acid reacts per mole of the base.

You must know the student will reach equivalence point when moles of lactic acid = moles NaOH.

the student will titrate the 0.126g of H₃C-CH(OH)-COOH. In moles (Using molar mass) are:

0.126g ₓ (1mol / 90.08g) = 1.40x10⁻³ moles of H₃C-CH(OH)-COOH

To reach equivalence point, the student must add 1.40x10⁻³ moles of NaOH. These moles comes from:

1.40x10⁻³ moles of NaOH ₓ (1L / 0.0500moles NaOH) = 0.0280L of the 0.0500M NaOH =

Answer:

Ksp = 2.06x10⁻¹⁰

Explanation:

For AB₂. solubility product constant, Ksp, is written as follows:

AB₂(s) ⇄ A²⁺ + 2Br⁻

Ksp = [A²⁺] [Br⁻]²

Molar solubility represents how many moles of AB₂ are soluble per liter of solution. and is obtained from Ksp:

AB₂(s) ⇄ A²⁺ + 2Br⁻

AB₂(s) ⇄ X + 2X

where X are moles that are soluble (Molar solubility)

Ksp = [X] [2X]²

As molar solubility of the salt is 3.72x10⁻⁴M:

Ksp = 4X³

Ksp = 4(3.72x10⁻⁴)³

Answer:

0.0668g

Explanation:

Step 1: Given data

Concentration of lead: 668 ppm (mg/kg)

Mass of water: 100.0 g

Step 2: Convert the mass of water to kilograms

We will use the relationship 1 kg = 1,000 g.

[tex]100.0g \times \frac{1kg}{1,000g} = 0.1000kg[/tex]

Step 3: Calculate the mass of lead in 0.1000 kg of water

There are 668 mg of Pb in 1 kg of water.

[tex]0.1000kgWater \times \frac{668mgPb}{1kgWater} = 66.8mgPb[/tex]

Step 4: Convert the mass of Pb to grams

We will use the relationship 1 g = 1,000 mg.

[tex]66.8mg \times \frac{1g}{1,000mg} = 0.0668g[/tex]

Answer:

1447584654 years or 1.44 billion years

Explanation:

From the formula;

0.693/t1/2=2.303/t log No/N

Where;

t1/2 = half life of the U-235 nuclides

No= amount of U-235 initially present

N= amount of U-235 present after a time t

t= time taken for N amount of U-235 to remain

Since N= 0.24No

Substituting values

0.693/703×10^6 = 2.303/t log No/0.24No

9.8578×10^-10 = 2.303/t log 1/0.24

9.8578×10^-10 = 1.427/t

t= 1.427/9.8578×10^-10

t= 1447584654 years

Answer:

Explanation:

The subscripts in a formula determine the ratio of the moles of each element in the compound. To convert this formula to the empirical formula, divide each subscript by 2. This is similar to reducing a fraction to its lowest denominator.

Answer:

Ph3OH

Explanation:

The reaction is between ethyl benzoate (PhCOOC2H5) and a Grignard reagent PhMgBr(C6H5MgBr).

The first step in the reaction mechanism is that the ethyl benzoate is converted to PhCOPh by the first molecule of Grignard reagent.

The second molecule of Grignard reagent now converts PhCOPh to Ph3O^-. In the presence of acid, Ph3O^- is now protonated to yield Ph3OH which is the major organic product of the reaction.

See image attached for more details.

Answer:

2NaOH(s) → Na₂O(s) + H₂O(g)

Hope that helps.

Answer:

Cr(OH)2(s), Na+(aq), and NO3−(aq)

Explanation:

Let is consider the molecular equation;

2NaOH(aq) + Cr(NO3)2(aq) -----> 2NaNO3(aq) + Cr(OH)2(s)

This is a double displacement or double replacement reaction. The reacting species exchange their partners. We can see here that both the sodium ion and chromium II ion both exchanged partners and picked up each others partners in the product.

Sodium ions and nitrate ions now remain in the solution while chromium II hydroxide which is insoluble is precipitated out of the solution as a solid hence the answer.

Answer:

pH = 10.87

Explanation:

You can find pH of a buffer (Mixture of a weak base, X, and its conjugate acid, HX) by using H-H equation:

pH = pKa + log [X] / [HX]

pKb is -log Kb = 3.13

Using Hint pKw = pKa + pKb (pKw = 14)

pKw = pKa + pKb

14 - 3.13 = pKa

10.87 = pKa

Replacing in H-H equation:

pH = 10.87 + log [0.42M] / [0.42]

pH = 10.87 + log 1

Answer:

k= 1.925×10^-4 s^-1

1.2 ×10^20 atoms/s

Explanation:

From the information provided;

t1/2=Half life= 1.00 hour or 3600 seconds

Then;

t1/2= 0.693/k

Where k= rate constant

k= 0.693/t1/2 = 0.693/3600

k= 1.925×10^-4 s^-1

Since 1 mole of the nuclide contains 6.02×10^23 atoms

Rate of decay= rate constant × number of atoms

Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms

Rate of decay= 1.2 ×10^20 atoms/s

Answer:

1-remain the same

2- remain the same

3-decrease

--------------------------

- decrease

- As the temperature is decreased, the reaction shifts from left to right. The product concentrations increase, and the reactant concentrations decrease. This corresponds to an increase in Kc.

--------------------------------

1- decrease

2-increase

3-decrease

4-remain the same

5-decrease

Explanation:

According to Le Chateliers principle, an increase in the volume of a gaseous system at equilibrium will shift the equilibrium position towards the side in which there are less volumes. Hence the answers written. When there is no change in volume, the number of moles of products remain the same.

-------------------------------

For an exothermic reaction, increasing the temperature shifts the equilibrium to the lefthand side.

When temperature is decreased, the equilibrium position will shift towards the right according to Le Chateliers principle

---------------------

Addition of a catalyst aids the reaction in which CO is consumed to proceed faster hence CO decreases in the system.

Since the reaction is exothermic, according to Le Chateliers principle, when the temperature is increased, the equilibrium position shifts towards the lefthand side and more CO is now present in the system.

When the pressure of the system is increased, the equilibrium position will shift towards the right hand side and more CO is converted to products hence its concentration in the system decreases.

Addition of argon gas has no effect on the equilibrium position since it does not participate in the reaction. However, addition of the reactant gases increases the rate of reaction and shifts the equilibrium position towards the right hand side thus decreasing the concentration of CO in the system.

Answer:

The correct answer is - alternating and direct, in order.

Explanation:

Alternating current is is type of electric current that is characterized by the direction of the flow of electrons in continuously switches its directs in opposite manner at regular cycles. While direct current or DC is flow of the electrons that move from starting to end in one direction.

Spinning turbines always leads to the alternating electric current while only solar energy produces the direct current with the help of the solar panels.

Thus, the correct answer is -  alternating and direct, in order.

Answer:

1. alternating

2. direct

3. The sun heats up the atmosphere as Earth spins, creating areas of high and low temperature. This temperature difference causes wind to start moving through convection, which can then drive a wind turbine to produce electricity.

Explanation:

From Penn

Answer:

The  wavelength is  [tex]\lambda = 91nm[/tex]  

Explanation:

From the question we are told that

    The energy required is  [tex]E = 13.6 \ eV[/tex]

This energy needed in form of a photon  can be mathematically represented as

         [tex]E = \frac{h * c }{\lambda }[/tex]

where h is the Planck constant with a value  [tex]h = 4.1357 * 10{-15} eV \cdot sec[/tex]

and  c is the speed of light which is  

 substituting values      

         [tex]13.6 eV = \frac{ 4.1357 * 10^{-15} eV * 3.0*10^8 }{\lambda }[/tex]

= >   [tex]\lambda = 9.1*10^{-8}[/tex]    

      [tex]\lambda = 91nm[/tex]  

Answer:

256 milimoles of Sodium Carbonate are in the flask

Explanation:

The chemist adds 180mL of a 1.42M sodium carbonate solution.

Molarity is an unit of concentration in chemistry defined as the ratio between moles of solute (Sodium Carbonate, in this case) per liter of solution.

A 1.42M solution contains 1.42 moles of Sodium carbonate in 1L of the solution.

As you have 180mL = 0.180L, moles of sodium carbonate you have are:

0.180L × (1.42 moles Sodium Carbonate / 1L) = 0.2556 moles of sodium carbonate

1000milimoles are equal to 1mole. 0.2556 moles are:

0.2556 moles × (1000 milimoles / 1 mole) =

Answer:

The change in internal energy of a system can be found by combining the heat energy added to a system minus the work done by the system is not a direct application of second law of thermodynamics.

Explanation:

Second law of thermodynamics states that heat can be transfer spontaneously from high temperature to low temperature only.

The change in internal energy of a system can be found by combining the heat energy added to a system minus the work done by the system is not a direct application of second law of thermodynamics because according to the second law of thermodynamics it is impossible in any system for heat transfer from a source to completely convert to work done in a cyclical process( bring the system to its original stage after each cycle) in which the system then return to it's original stage.

Answer:the mass is 35456

Explanation:

because it works

The density has been the division of the mass to the volume of the substance. The mass of an iron cannonball of density 7.86 grams/cm³ is estimated to be 0.951 kilograms.

The mass (m) has been calculated by multiplying the density (D or ρ) and volume (V) of the iron cannonball. The measurement of the amount is called mass whereas the measurement of the space occupied by the substance volume is given as density.

The amount of the substance contained in a volume (milliliters, liters, and cubic centimeters) along with the density of the substance has been used to give mass. The mass can be estimated in kilograms, grams, milligrams, etc.

Given,

Density of iron cannonball (D) = 7.86 grams/cm³

Volume of cannonball (V) = 121 cm³

The formula to calculate mass is given as:

Density = mass ÷ volume

Mass = density × volume

Substituting values in the above formula, the mass of the iron cannonball in grams is calculated as:

mass = 7.86 grams/cm³ × 121 cm³

mass = 951.06 grams

Therefore, the iron cannonball with a density of 7.86 grams/cm³ has a mass of 0.951 kilograms.

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Your question is incomplete, but most probably your full question was, A cannonball made of iron has a volume of 121 cm³. if the iron has a density of 7.86 g/cm³, what is the mass, in kilograms, of the cannonball?

Answer:

Polar molecules are not symmetrical

Explanation:

Even though the structures of the molecules involved were not shown in the question, but I will proceed to give a general explanation of the conditions that describe a polar molecule.

First of all, symmetrical molecules are non-polar and asymmetrical molecules are polar. This is the reason why CF4 will be a nonpolar molecule but H2O will be a polar molecule. Some symmetrical molecules may may posses polar bonds or dipoles but these dipoles eventually cancel out since the molecule is symmetrical in nature.

Summarily, if a molecule possess the same type of atoms attached to the central atom with some symmetry axes, like C3, C4 etc., we will end up with a non polar molecule but if we have a nonplanar molecule, then we will end up finding it to be polar.

Answer:

[tex]P_{gas}=131.96torr[/tex]

Explanation:

Helo,

In this case, the pressure must be computed as follows:

[tex]P_{gas}=gh\rho _{Hg}[/tex]

Which is using the density of mercury (13.6 g/mL) and its height, thus, we obtain (using the proper units):

[tex]P_{gas}=9.8\frac{m}{s^2}*(13.2cm *\frac{1m}{100cm} )*(13.6\frac{g}{cm^3}*\frac{1kg}{1000g}*\frac{1x10^6cm^3}{1m^3} )\\\\P_{gas}=17592.96Pa*\frac{760torr}{101325Pa} \\\\P_{gas}=131.96torr[/tex]

Best regards.

The correct answer is C. There is no oxygen in space, so there can be no combustion.

Explanation:

Fire and flames are the result of a chemical process known as combustion. Moreover, for combustion to occur there are two essential elements. The first one is a fuel or a substance that releases energy and ignites, and the second one is an oxidant, which accepts electrons. This mix and reaction causes high temperatures and release of heat in the form of fire and flames.

This implies, that for fireballs or any other form of fire to exist there must be oxygen or any substance that replaces it. This does not occur in space because the levels of oxygen are extremely low, this means, at least oxygen is added fireballs are not possible in this context as there is no oxygen, and therefore no combustion (Option C).

Answer:

C.) There is no oxygen in space, so there can be no combustion.

Explanation:

I got it correct on founders edtell

Answer:

1. False

2. True

3. True

4. True

5. True

6. True

7. True

8. False

9. True

Explanation:

Density is a physical property since its measurement does not involve any chemical process.

Since transition elements exhibit variable oxidation states, the actual oxidation state of the transition element must be specified in the compound.

Due to the fact that neutron has no charge, it was discovered by Chadwick long after the electron and proton were discovered.

The balancing of chemical reaction equations is a demonstration that atoms are neither created no destroyed. It also shows that mass is neither created nor destroyed in chemical reactions.

When a gas is heated, it expands. Its volume and its kinetic energy increases. Since volume and pressure are inversely proportional (Boyle's law) the pressure decreases.

Enthalpy is said to be an extensive property. This implies that the magnitude of change in enthalpy is known to depend on the amount of reactants that is actually reacted.

The combined gas law is applicable to all ideal gas systems irrespective of their individual chemical formulas.

10g of CO2 contains 0.227 moles of CO2 while 10g of NO contains 0.33 moles of NO hence 10.0 g of NO will contain more molecules than 10.0g of CO2.

If a sample is not at absolute zero, the particles are known to possess kinetic energy which decreases continuously until absolute zero is attained.

Answer:

The side chains of the amino acids alternate above and below the sheet

Explanation:

Hydrogen bonds are formed between the amine and carbonyl groups across strands. ... Antiparallel ß sheets are slightly more stable than parallel ß sheets because the hydrogen bonding pattern is more optimal.

Answer:

The correct answer is d. Sn(s) | Sn²⁺(aq) ∥ H⁺(aq) | H₂(g) | Pt

Explanation:

The half reactions are:

2H⁺(aq) + 2 e- ⟶ H₂(g) (reduction)

Sn(s) ⟶ Sn²⁺(aq) + 2 e-  (oxidation)

In the cell notation, there are two electrodes in which are separated the reduction reaction from the oxidation reaction. In the left electrode occurs the oxidation reaction (anode) while in the right electrode occurs the reduction reaction (cathode). The general form of the cell notation is the following:

anode reaction∥ cathode reaction

where the two bars ( ∥ ) represent the physical barrier between the electrodes. A single bar ( | ) is used to represent a phase separation.  

In this redox reaction, the half reaction of the anode is Sn(s) ⟶ Sn²⁺(aq) + 2 e-; whereas the half reaction of the cathode is 2H⁺(aq) + 2 e- ⟶ H₂(g).

The componens are written in order according to the half reaction. Since Sn²⁺ and H⁺ ions are in solution, a platinum electrode is used and represented as Pt. Thus, the cell notation is:

Sn(s) | Sn²⁺(aq) ∥ H⁺(aq) | H₂(g) | Pt