How many molecules are in 23 Moles of oxygen o2

4.3 moles of a gas are at a temperature of 28 degrees * C with a pressure of 1.631 atm. What volume does the gas occupy?

Shkiper50 [21]

Answer:

65.2L

Explanation:

Using the general gas equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (Litres)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (Kelvin)

According to the information provided in this question,

P = 1.631 atm

V = ?

n = 4.3 moles

T = 28°C = 28 + 273 = 301K

Using PV = nRT

V = nRT/P

V = 4.3 × 0.0821 × 301 ÷ 1.631

V = 106.26 ÷ 1.631

V = 65.15

Volume of the gas = 65.2L

7 0

3 years ago

A reaction was performed in which 3.6 g 3.6 g of benzoic acid was reacted with excess methanol to make 1.3 g 1.3 g of methyl ben

Anit [1.1K]

Answer: The percent yield of the reaction is 32.34 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For benzoic acid:

Given mass of benzoic acid = 3.6 g

Molar mass of benzoic acid = 122.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of benzoic acid}=\frac{3.6g}{122.12g/mol}=0.0295mol$

The chemical equation for the reaction of benzoic acid and methanol is:

$\text{Benzoic acid + methanol}\rightarrow \text{methyl benzoate}$

By Stoichiometry of the reaction

1 mole of benzoic acid produces 1 mole of methyl benzoate

So, 0.0295 moles of benzoic acid will produce = $\frac{1}{1}\times 0.0295=0.108$ moles of methyl benzoate

Now, calculating the mass of methyl benzoate from equation 1, we get:

Molar mass of methyl benzoate = 136.15 g/mol

Moles of methyl benzoate = 0.0295 moles

Putting values in equation 1, we get:

$0.0295mol=\frac{\text{Mass of methyl benzoate}}{136.15g/mol}\\\\\text{Mass of methyl benzoate}=(0.0295mol\times 136.15g/mol)=4.02g$

To calculate the percentage yield of methyl benzoate, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of methyl benzoate = 1.3 g

Theoretical yield of methyl benzoate = 4.02 g

Putting values in above equation, we get:

$\%\text{ yield of methyl benzoate}=\frac{1.3g}{4.02g}\times 100\\\\\% \text{yield of methyl benzoate}=32.34\%$

Hence, the percent yield of the reaction is 32.34 %

6 0

3 years ago

What are the answers please help

LenaWriter [7]

Thank you for the free 15 points .

4 0

2 years ago

List three factors thatinfluence the rate at whick dissolving occurs

lys-0071 [83]

1) Temperature (heat) of the solution
2) Concentration (amount) of both solvent (usually water) and solute (substance being dissolved by solvent)
3) Movement (kinetic energy) of the solution, as in shaking/stirring

5 0

3 years ago

Which of the followings is true about G0'? A. G0' can be determined using Keq' B. G0' indicates if a reaction can occur under no

nekit [7.7K]

Answer:

A and D are true , while B and F statements are false.

Explanation:

A) True. Since the standard gibbs free energy is

ΔG = ΔG⁰ + RT*ln Q

where Q= [P1]ᵃ.../([R1]ᵇ...) , representing the ratio of the product of concentration of chemical reaction products P and the product of concentration of chemical reaction reactants R

when the system reaches equilibrium ΔG=0 and Q=Keq

0 = ΔG⁰ + RT*ln Q → ΔG⁰ = (-RT*ln Keq)

therefore the first equation also can be expressed as

ΔG = RT*ln (Q/Keq)

thus the standard gibbs free energy can be determined using Keq

B) False. ΔG⁰ represents the change of free energy under standard conditions . Nevertheless , it will give us a clue about the ΔG around the standard conditions .For example if ΔG⁰>>0 then is likely that ΔG>0 ( from the first equation) if the temperature or concentration changes are not very distant from the standard conditions

C) False. From the equation presented

ΔG⁰ = (-RT*ln Keq)

ΔG⁰>0 if Keq<1 and ΔG⁰<0 if Keq>1

for example, for a reversible reaction ΔG⁰ will be <0 for forward or reverse reaction and the ΔG⁰ will be >0 for the other one ( reverse or forward reaction)

D) True. Standard conditions refer to

T= 298 K

pH= 7

P= 1 atm

C= 1 M for all reactants

Water = 55.6 M

5 0

3 years ago