Balance equation for 15 g of solid Mg reacts with 15 g of HCl and produce MgCl2 and H2

To solve the financial dilemma, the chemical company can consider using the excess silver nitrate solution to synthesize silver nanoparticles, which have various applications in industries.

Silver nanoparticles can be synthesized by reducing silver ions with a reducing agent, and silver nitrate can serve as a source of silver ions.

One possible reaction for the synthesis of silver nanoparticles using silver nitrate is the reduction of silver ions with sodium borohydride (NaBH4). The reaction can be represented as:

AgNO3 + NaBH4 → Ag nanoparticles + NaNO3 + B2H6

In this reaction, silver nitrate is the oxidizing agent, which accepts electrons, while sodium borohydride is the reducing agent, which donates electrons to reduce the silver ions. The reaction also produces sodium nitrate and borane gas as byproducts.

The synthesized silver nanoparticles can be sold to various industries, generating revenue for the company and potentially reducing their debt. The company can also consider optimizing the synthesis process to increase the yield and purity of the silver nanoparticles, which can increase their market value.

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Ionization energy for the neutral Li atom in its ground state is approximately 520.9 kJ/mol.

The energy required to remove an electron from an atom in its ground state is the ionization energy. In this problem, we are given the wavelengths of various transitions of neutral Li atoms. From these wavelengths, we can calculate the energy of each transition using the equation E=hc/λ,

where h is Planck's constant,

c is the speed of light

λ is the wavelength.

Using the given wavelengths, we can calculate the energy of each transition and determine the difference in energy between the ground state and the excited state. The ionization energy is the energy required to remove an electron from the ground state, which is equal to the energy difference between the ground state and the ionized state.

In this case, the transition from the ground-state configuration 1s²2s¹ to the 2P term occurs at 670.78 nm. From this, we can calculate the energy difference between the ground state and the 2P term. Then, by adding the energy differences between the 2P and 2D terms, and the 2D and 2S terms, we can calculate the ionization energy of the ground-state atom. As a result, when the temperature lowers to 8.5°C in the evening, the volume of the vessel is roughly 2.64 L.

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The primary alcohol required for the synthesis of heptanoic acid is heptanol, which has the chemical formula C₇H₁₆O. The aldehyde required for this synthesis is heptanal, which has the chemical formula C₇H₁₄O.

Heptanoic acid is a carboxylic acid with seven carbon atoms. It can be synthesized from primary alcohol and an aldehyde via oxidation.

To synthesize heptanoic acid, heptanol, and heptanal are reacted in the presence of an oxidizing agent, such as potassium permanganate (KMnO₄) or chromium trioxide (CrO₃). The oxidation of heptanol produces heptanal, which is further oxidized to heptanoic acid. The chemical equation for the synthesis of heptanoic acid is as follows:

C₇H₁₆O + O → C₇H₁₄O + H₂O

C₇H₁₄O + O → C₇H₁₂O₂ + H₂O

The resulting product, heptanoic acid, is a colorless liquid with a pungent odor and is commonly used as a flavoring agent and in the production of esters for fragrances and plastics.

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The pressure will be 139.92 kPa at a volume of 27.5 mL.

To answer this question, we will use Boyle's Law formula, which states that the product of the initial pressure (P1) and volume (V1) of a gas is equal to the product of the final pressure (P2) and volume (V2) when the temperature remains constant.


Step 1: Identify the initial pressure (P1), initial volume (V1), and final volume (V2).

P1 = 102.3 kPa
V1 = 37.5 mL
V2 = 27.5 mL

Step 2: Apply Boyle's Law formula, which is P1 * V1 = P2 * V2. We need to find the final pressure (P2).

102.3 kPa * 37.5 mL = P2 * 27.5 mL

Step 3: Solve for P2.

P2 = (102.3 kPa * 37.5 mL) / 27.5 mL

Step 4: Calculate the value of P2.

P2 ≈ 139.64 kPa

At 27.5 mL, the pressure of the gas will be approximately 139.64 kPa.

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The mass of dissolved solids in 1 m³ of the water sample is 16 g.

To convert from cm³ to m³, we divide by 1,000,000 (10^6) since there are 1,000,000 cm³ in 1 m³.

First, we need to find the mass of dissolved solids in 1 cm³ of the water sample:

Next, we can find the mass of dissolved solids in 1 m³ of the water sample:

However, the answer should be given in standard form, so we convert 640 to scientific notation:

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The final pressure inside the tank is 88.9 kPa.

To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas.

The combined gas law is given by:

(P1 * V1) / (T1) = (P2 * V2) / (T2)

where

P1 and T1 are the initial pressure and temperature of the gas,

V1 is the initial volume of the gas,

P2 is the final pressure of the gas,

V2 is the final volume of the gas, and

T2 is the final temperature of the gas.

We can assume that the volume of the gas in the tank remains constant, since it is a steel tank. Therefore, V1 = V2.

We can convert the temperatures to Kelvin by adding 273.15 to each temperature value. Therefore,

T1 = 173.15 K and

T2 = 308.15 K.

Substituting these values into the combined gas law, we get:

(50.0 kPa * V1) / (173.15 K) = (P2 * V1) / (308.15 K)

P2 = (50.0 kPa * 308.15 K) / 173.15 K

P2 = 88.98 kPa

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Answer:

88.98 kPa (2 d.p.)

Explanation:

To find the final pressure inside the steel tank, we can use Gay-Lussac's law since the volume is constant.

[tex]\boxed{\sf \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}}[/tex]

where:

As we are solving for the final pressure, rearrange the equation to isolate P₂:

[tex]\sf P_2=\dfrac{P_1T_2}{T_1}[/tex]

Convert the given temperatures from Celsius to Kelvin by adding 273.15:

[tex]\implies \sf T_1=-100+273.15=173.15\;K[/tex]

[tex]\implies \sf T_2=35+273.15=308.15\;K[/tex]

Therefore, the values to substitute into the equation are:

Substitute the values into the equation and solve for P₂:

[tex]\implies \sf P_2=\dfrac{50.0\cdot 308.15}{173.15}[/tex]

[tex]\implies \sf P_2=\dfrac{15407.5}{173.15}[/tex]

[tex]\implies \sf P_2=88.98354028...[/tex]

[tex]\implies \sf P_2=88.98\;kPa\;(2\;d.p.)[/tex]

Therefore, the final pressure inside the steel tank is 88.98 kPa when the temperature is increased from -100.0°C to 35.0°C.

Answer: polypeptide chain

Explanation:

The correct answer is therefore A, -1.66 V.

The given information includes the standard reduction potential of the half-reaction:

Ag(aq) + e- → Ag(s) E° = +0.80 V

We can use this information along with the standard cell potential equation to find the standard reduction potential of the half-reaction:

E°cell = E°reduction + E°oxidation

where E°cell is the standard cell potential, E°reduction is the reduction potential for the half-reaction being reduced, and E°oxidation is the oxidation potential for the half-reaction being oxidized.

In this case, the two half-reactions involved are:

M3+(aq) + 3e- → M(s) (reduction)

3Ag(aq) → 3Ag+(aq) + 3e- (oxidation)

The reduction half-reaction needs to be flipped and its potential sign changed to obtain the oxidation potential:

M(s) → M3+(aq) + 3e- (oxidation)

The standard cell potential is the difference between the reduction and oxidation potentials:

E°cell = E°reduction + E°oxidation

E°cell = E°(M3+(aq) + 3e- → M(s)) + E°(M(s) → M3+(aq) + 3e-)

E°cell = E°(M(s) → M3+(aq) + 3e-) + (-E°(3Ag(aq) → 3Ag+(aq) + 3e-))

E°cell = E°(M(s) → M3+(aq) + 3e-) - E°(Ag(aq) → Ag(s))

E°cell = E°(M3+(aq) + 3e- → M(s)) - E°(Ag(aq) → Ag(s))

E°cell = -2.46 V - 0.80 V = -3.26 V

Therefore, the standard reduction potential for the half-reaction M3+(aq) + 3e- → M(s) is:

E°(M3+(aq) + 3e- → M(s)) = E°cell + E°(Ag(aq) → Ag(s))

E°(M3+(aq) + 3e- → M(s)) = -3.26 V + 0.80 V = -2.46 V

The correct answer is therefore A, -1.66 V.

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To solve this problem, we can use the formula:

q = m*c*ΔT, where q is the amount of heat energy absorbed by the gold, m is the mass of the gold, c is the specific heat capacity of gold, and ΔT is the change in temperature of the gold.

We are given the mass of gold (m = 4.1 g), the specific heat capacity of gold (c = 0.130 J/g °C), and the amount of energy used to heat the gold (q = 52.2 J). We are asked to find the final temperature of the gold (ΔT).

Rearranging the formula, we get:
ΔT = q/(m*c)
Substituting the values we know, we get:
ΔT = 52.2 J / (4.1 g * 0.130 J/g °C)
ΔT = 98.92 °C
This is the change in temperature of the gold. To find the final temperature, we add this to the original temperature of 25.0°C:
Final temperature = 25.0°C + 98.92°C
Final temperature = 123.92°C
Therefore, the final temperature of the gold is 123.1°C.

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Answer:

1000m/s^2

Explanation:

since F=ma

:a=m/F

Yo convert ur m from kg to g

20×1000

20000

a=20000\20

a=1000m/s^2

When 250.0 g of lead is cooled from 15.0°C to 12.0°C, approximately 29.36 calories of energy (heat) are released.

To calculate the amount of energy (heat) released when 250.0 g of lead is cooled from 15.0°C to 12.0°C, we need to use the specific heat capacity and the change in temperature of lead. The specific heat capacity of lead is 0.128 J/g°C.

The equation to calculate the energy released (Q) is:

Q = m × c × ΔT

where m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

Substituting the given values, we get:

Q = 250.0 g × 0.128 J/g°C × (12.0°C - 15.0°C)

Q = - 122.88 J

The negative sign indicates that energy is being released, as the lead is losing heat. To convert this to calories, we need to divide by the conversion factor of 4.184 J/cal:

Q = - 122.88 J ÷ 4.184 J/cal

Q ≈ - 29.36 cal

Therefore, when 250.0 g of lead is cooled from 15.0°C to 12.0°C, approximately 29.36 calories of energy (heat) are released.

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Manganese(IV) permanganate is a chemical compound with the formula [tex]MnO4[/tex]. It is an ionic compound that consists of one manganese atom and four oxygen atoms.

The oxidation state of manganese in the compound is +7, which means that it has lost seven electrons and has seven fewer electrons than the neutral atom. The oxidation state of oxygen in the compound is -2, which means that each oxygen atom has gained two electrons.

To calculate the number of moles of oxygen in one mole of manganese(IV) permanganate, we can use the molecular formula of the compound, which tells us that there are four oxygen atoms per one manganese atom. Therefore, the molar ratio of oxygen to manganese is 4:1.

So, one mole of manganese(IV) permanganate contains four moles of oxygen. This can be written as:

1 mole [tex]MnO4[/tex] = 4 moles O2

This means that if we have one mole of manganese(IV) permanganate, we would have four moles of oxygen atoms.

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The molar mass of [tex]CHOCH[/tex] is 64.05 g/mol, which means that one mole of [tex]CHOCH[/tex] has a mass of 64.05 grams.

The molar mass of a compound is the mass in grams of one mole of the substance. To calculate the molar mass of [tex]CHOCH[/tex], we need to determine the atomic masses of all the atoms in one molecule of the compound and add them together.

[tex]CHOCH[/tex] has one carbon (C) atom, three oxygen (O) atoms, and four hydrogen (H) atoms. The atomic mass of C is 12.01 g/mol, O is 16.00 g/mol, and H is 1.01 g/mol. Therefore, we can calculate the molar mass of [tex]CHOCH[/tex] as follows:

Molar mass = (1 x atomic mass of C) + (3 x atomic mass of O) + (4 x atomic mass of H)

Molar mass = (1 x 12.01) + (3 x 16.00) + (4 x 1.01)

Molar mass = 64.05 g/mol

Therefore, the molar mass of [tex]CHOCH[/tex] is 64.05 g/mol, which means that one mole of [tex]CHOCH[/tex] has a mass of 64.05 grams.

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The substance described in the question is most likely a base or an alkali. Bases have a bitter taste, feel slippery or soapy to the touch, conduct electricity in solution, and have a pH above 7.

The slipperiness is due to the ability of bases to react with oils and fats to form soaps, which have a slippery texture.

The ability to conduct electricity is due to the presence of ions in the solution. In the case of bases, these are usually hydroxide ions (OH-) which can conduct electric current when dissolved in water.

The high pH is also characteristic of bases, as pH is a measure of the concentration of hydrogen ions (H+) in solution. In the case of bases, the concentration of OH- ions is higher than the concentration of H+ ions, leading to a pH above 7.

Examples of common bases include "sodium hydroxide (NaOH), potassium hydroxide (KOH), and calcium hydroxide (Ca(OH)2)".

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Answer: B. Energy

Explanation: The matter is converted into energy, which is released in the form of radiation, this is due to the fact that the mass of the products of the reaction is less than the mass of the reactants, and this difference in mass is converted into energy. Aka ([tex]E=mc^{2}[/tex]).

The pyramid shape is a good way to model the relative amount of energy in different groups of organisms in a food chain because it reflects the energy transfer from one trophic level to another.

In a food chain, energy is transferred from one organism to another through the consumption of food. As each organism consumes the one below it, a large proportion of the energy that was stored in the previous organism is lost as heat or used for metabolic processes such as respiration. This means that there is less energy available for the next organism in the chain.

The pyramid shape reflects this decrease in available energy at each trophic level. The base of the pyramid represents the primary producers, which have the largest amount of energy available to them through photosynthesis. As we move up the pyramid to the next trophic level, the available energy decreases, representing the loss of energy as we move up the food chain.

By using a pyramid shape to model the relative amount of energy in different groups of organisms in a food chain, we can see the significant decrease in available energy at each successive trophic level. This shape helps to illustrate the importance of primary producers in supporting life on Earth and the delicate balance of energy transfer that exists in ecosystems.

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A chemist interested in the efficiency of a chemical reaction would calculate the percent yield. To do this, follow these steps:

1. Determine the balanced chemical equation for the reaction, which shows the stoichiometric relationship between reactants and products.

2. Identify the limiting reactant by comparing the initial amounts of reactants to their stoichiometric ratios in the balanced equation.

3. Calculate the theoretical yield by using the stoichiometric relationship between the limiting reactant and the desired product, based on their balanced chemical equation.

4. Measure the actual yield of the product obtained from the experiment.

5. Calculate the percent yield using the formula: (Actual yield / Theoretical yield) × 100%.

This process will provide the chemist with a measure of the efficiency of the chemical reaction.

Complete question : A chemist interested in the efficiency of a chemical reaction would calculate the percent yield ?

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Answer:

It off soudium and i know this from experments so the answear is b

Explanation:

A 35. 3 g of element M is reacted with nitrogen to produce 43. 5 g of compo und M₃N₂ (i) The molar mass of element M is 24.0 g/mol. (ii) The name of the element is magnesium (Mg).

(i) To find the molar mass of element M, we need to use stoichiometry to relate the mass of M to the mass of M₃N₂. We can start by calculating the moles of M3N2 produced:

43.5 g M₃N₂ × 1 mol M₃N₂/100.9 g M₃N₂ = 0.43 mol M₃N₂

Since the molar ratio between M and M₃N₂ is 1:3, we can calculate the moles of M:

0.43 mol M₃N₂ × 1 mol M/3 mol M₃N₂ = 0.14 mol M

Finally, we can calculate the molar mass of M by dividing its mass (35.3 g) by the number of moles (0.14 mol):

molar mass of M = 35.3 g/0.14 mol = 253 g/mol

However, this value is much higher than the molar mass of any known element. We can recognize that the formula M₃N₂ implies that element M is a metal with a +2 charge, since each M atom forms 3 bonds with N atoms, and each N atom forms 2 bonds with M atoms.

(ii) Based on this information, we can identify element M as magnesium (Mg), which has a molar mass of 24.0 g/mol.

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To manipulate seven additional data sets and place these values in your Ocean Interactions, follow these steps:
Step 1: Obtain the data sets
First, acquire the seven additional data sets that you want to include in your Ocean Interactions analysis. These data sets could be related to variables such as temperature, salinity, ocean currents, or marine life distributions.

Step 2: Organize the data
Next, organize the data sets by sorting, filtering, or aggregating them as needed to make them more manageable for analysis. This process may involve cleaning the data to remove any inconsistencies or errors, as well as converting the data into a compatible format for further manipulation.

Step 3: Manipulate the data
Using various data manipulation techniques, transform the additional data sets to create new variables or features that can help provide a deeper understanding of the Ocean Interactions. This manipulation could include calculations, statistical analysis, or creating visual representations to identify patterns or trends within the data.

Step 4: Integrate the data
Combine the manipulated additional data sets with the existing Ocean Interactions data to create a comprehensive analysis. This integration process may involve merging or joining data sets based on common variables or geographical locations, ensuring that the resulting data accurately reflects the interactions between various ocean-related factors.

Step 5: Analyze the data
With the additional data sets now integrated into your Ocean Interactions analysis, examine the relationships between the different variables to gain insights into the complex dynamics at play. This analysis could involve statistical tests, correlations, or predictive modeling techniques to better understand the underlying patterns and trends in the data.

Step 6: Interpret the results
Based on the analysis, draw conclusions about the role of the additional data sets in the overall Ocean Interactions. This interpretation should consider the potential implications of these findings for the broader understanding of ocean processes and the management of marine ecosystems.

By following these steps, you will successfully manipulate seven additional data sets and place these values in your Ocean Interactions analysis, enhancing your understanding of the complex dynamics involved in the marine environment.

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Here, each of the elements below with the class to which it belongs.  

Lithium → Alkali metals

Uranium → Transition metals

What is an Alkali metals?

Alkali metals are a group of highly reactive chemical elements in the periodic table. These elements include lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr). Alkali metals have a single electron in their outermost shell, which makes them highly reactive and able to easily lose that electron to form a positive ion. They are typically soft, silvery-white metals that have low melting and boiling points, and are highly reactive with water and other substances. Alkali metals are important in various industrial applications, such as batteries, alloys, and chemical synthesis.

Krypton → Noble gases

Manganese → Transition metals

Fluorine → Halogens

Barium → Alkaline Earth

Most reactive metal → Alkali metals

Silicon → Metalloids

Groups 3-12 → Transition metals

Most reactive nonmetals → Halogens

Inert and unreactive → Noble gases

Has characteristics of metals and nonmetals → Metalloids

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We can use the formula for calculating heat:

Q = m × c × ΔT

where Q is the amount of heat transferred, m is the mass of the substance, c is its specific heat, and ΔT is the change in temperature.

Plugging in the given values, we get:

8250 J = m × 0.9025 J/g ·°C × 40.0 °C

Simplifying, we get:

8250 J = m × 36.1 J/g

Solving for m, we get:

m = 8250 J ÷ 36.1 J/g

m ≈ 228.26 g

Therefore, the mass of the aluminum is approximately 228.26 g.

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The pH of the buffered solution is 4.374.

A buffered solution is a solution that resists changes in pH when small amounts of acid or base are added to it.

In order to calculate the pH of a buffered solution, we need to use the Henderson-Hasselbalch equation, which is pH = pKa + log([A-]/[HA]). In this equation, pKa is the dissociation constant of the weak acid (benzoic acid in this case), [A-] is the concentration of the conjugate base (sodium benzoate), and [HA] is the concentration of the weak acid.

First, we need to find the concentrations of benzoic acid and sodium benzoate in the solution. We can use the equation n = cV, where n is the number of moles, c is the concentration, and V is the volume.

For benzoic acid:
n = (21.5 g / 122.12 g/mol) = 0.176 mol
c = 0.176 mol / 0.2 L = 0.88 M
[HA] = 0.88 M

For sodium benzoate:
n = (37.7 g / 144.11 g/mol) = 0.262 mol
c = 0.262 mol / 0.2 L = 1.31 M
[A-] = 1.31 M

Next, we need to find the pKa of benzoic acid. The pKa of benzoic acid is 4.20.

Now we can plug in the values into the Henderson-Hasselbalch equation:
pH = 4.20 + log([1.31]/[0.88])
pH = 4.20 + log(1.49)
pH = 4.20 + 0.174
pH = 4.374

Therefore, the pH of the buffered solution is 4.374.

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We can distinguish 4-hydroxy-4-methyl-2-pentanone (side product) from the major product.

To distinguish 4-hydroxy-4-methyl-2-pentanone (side product) from the major product using physical properties, you should consider the following factors:

1. Molecular weight: Calculate the molecular weight of both the side product and the major product. Different molecular weights result in different physical properties.

2. Boiling point: The boiling point of each compound may vary, so compare their boiling points to distinguish between them.

3. Melting point: Like the boiling point, the melting point of each compound may be different, allowing you to differentiate them.

4. Solubility: Check the solubility of both compounds in different solvents. Their solubility may differ in various solvents, helping you identify each compound.

5. Polarity: Determine the polarity of each compound by looking at their molecular structures. Different polarities can affect various physical properties, such as solubility and boiling points.

6. Spectroscopy: Analyze the compounds using spectroscopic techniques like infrared (IR), nuclear magnetic resonance (NMR), or mass spectrometry (MS). Each compound will have unique spectroscopic properties, which can be used for identification.

By comparing and analyzing these physical properties, you can distinguish 4-hydroxy-4-methyl-2-pentanone (side product) from the major product.

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Yes, two magnets can create magnetic force fields that allows them to interact without touching.

Magnetic forces are non contact forces; they pull or push on objects without touching them. Magnets are only attracted to a few 'magnetic' metals and not all matter. Yes, the investigation did answer the question about whether two magnets create magnetic force fields that allow them to interact without touching.

The investigation provided enough evidence to support the idea that invisible magnetic force fields exist:

The investigation involved observing how two magnets interact with each other without touching. The magnets were brought closer together until they interacted, and then they were moved further apart. This process was repeated several times, and the results were observed and recorded. During the investigation, it was observed that the magnets interacted with each other even when they were not touching. This interaction occurred because the magnets created magnetic force fields that allowed them to interact with each other even when they were not in direct contact. This is because the interaction between the magnets could not be explained by any other means except through the existence of magnetic force fields.

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The concept Boyle's law is used here to determine the new pressure of air inside the balloon. For a gas the relationship between volume and pressure is expressed using Boyle's law. The new pressure is 400 kPa.

The Boyle's law states that at constant temperature, the volume of a given mass of gas is inversely proportional to its pressure. The product of pressure and volume of a given mass of gas is constant.

Mathematically PV = k

P₁V₁ = P₂V₂

P₂ = P₁V₁ / V₂

100 × 4 / 1 = 400 kPa

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The percent composition of the sample containing 1.29 grams of carbon and 1.71 grams of oxygen is 43% carbon and 57% oxygen.

The percent composition of a sample can be calculated by dividing the mass of each element in the sample by the total mass of the sample and then multiplying by 100%.

To find the percent composition of a sample containing 1.29 grams of carbon and 1.71 grams of oxygen, we need to calculate the total mass of the sample first.

Total mass of the sample = mass of carbon + mass of oxygen
= 1.29 grams + 1.71 grams
= 3 grams

Now, we can calculate the percent composition of carbon and oxygen in the sample:

Percent composition of oxygen = (mass of oxygen / total mass of the sample) x 100%
= (1.71 grams / 3 grams) x 100%
= 57%

Percent composition of carbon =  (mass of carbon / total mass of the sample) x 100%

=(1.29 grams / 3 grams) x 100%

= 43%

Therefore, the sample contains 43% carbon and 57% oxygen by mass.

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From the calculations, we can see that the mass of the acetic acid that is produced is 28.2 g.

In a chemical reaction involving two or more reactants, the limiting reactant is the reactant that is consumed completely, thereby limiting the amount of product that can be formed. The other reactant(s) that remain after the limiting reactant is completely consumed are called excess reactants.

Number of moles of CH3CHO = 20.8g/44 g/mol

= 0.47 moles

Number of moles of O2 = 14.5 g/32 g/mol

= 0.45 moles

If 2 moles of CH3CHO reacts with 1 mole of O2

0.47 moles of CH3CHO would react with 0.47 * 1/2

= 0.24 moles

Thus CH3CHO is the limiting reactant

Mass of the acetic acid produced = 0.47 moles * 60 g/mol

= 28.2 g

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Answer:

Molecular weight of CH4 is 16 CH4 has four hydrogen atoms 1 mole of a compound contain 6.023*1023 atoms 12 gm of CH4 = 0

When hydrogen and steam are both present in a gas at the same pressure and temperature, this is the ideal gas condition. This is so because according to the ideal gas law, an ideal gas's pressure, volume, and temperature are all precisely proportional to one another.

This indicates that when the two gases have the same temperature and pressure, the two gases will also have the same volume. As a result, the gases are in their ideal state, having the same volume and pressure but retaining their distinct chemical compositions.

This is perfect because it enables the two gases to interact with one another in a predictable way, allowing for the measurement and prediction of the gases' behaviour.

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