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2 years ago

Estimate ΔG°rxn for the following reaction at 449.0 K. CH2O(g) + 2 H2(g) → CH4(g) + H2O(g) ΔH°= -94.9 kJ; ΔS°= -224.2 J/K

Chemistry

1 answer:

Sliva [168]2 years ago

7 0

This equation relates all three variables, so just plug in all of the given values. Don’t forget to convert deltaS into kJ/K because Gibbs free energy is measured in kJ. The K used in temperature cancels out the K in kJ/K, leaving only kJ as your units. Don’t forget sig figs!

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A scuba tank contains a mixture of oxygen (O2) and nitrogen (N2) gas. The oxygen has a partial pressure of PO2=5.62MPa. The tota

dmitriy555 [2]

Answer:

21.16 MPa

Explanation:

Partial pressure of oxygen = 5.62 MPa

Total gas pressure = 26.78 MPa

But

Total pressure of the gas= sum of partial pressures of all the constituent gases in the system.

This implies that;

Total pressure of the system = partial pressure of nitrogen + partial pressure of oxygen

Hence partial pressure of nitrogen=

Total pressure of the system - partial pressure of oxygen

Therefore;

Partial pressure of nitrogen= 26.78 - 5.62

Partial pressure of nitrogen = 21.16 MPa

7 0

2 years ago

Read 2 more answers

How many grams of sodium acetate ( molar mass = 83.06 g/mol ) must be added to 1.00 Liter of a 0.200 M acetic acid solution to m

Pie

Answer: The mass of sodium acetate that must be added is 30.23 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of acetic acid solution = 0.200 M

Volume of solution = 1 L

Putting values in above equation, we get:

$0.200M=\frac{\text{Moles of acetic acid}}{1L}\\\\\text{Moles of acetic acid}=(0.200mol/L\times 1L)=0.200mol$

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[\text{acid}]})$

$pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of acetic acid = 4.74

$[CH_3COONa]=?mol$

$[CH_3COOH]=0.200mol$

pH = 5.00

Putting values in above equation, we get:

$5=4.74+\log(\frac{[CH_3COONa]}{0.200})$

$[CH_3COONa]=0.364mol$

To calculate the mass of sodium acetate for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of sodium acetate = 83.06 g/mol

Moles of sodium acetate = 0.364 moles

Putting values in above equation, we get:

$0.364mol=\frac{\text{Mass of sodium acetate}}{83.06g/mol}\\\\\text{Mass of sodium acetate}=(0.364mol\times 83.06g/mol)=30.23g$

Hence, the mass of sodium acetate that must be added is 30.23 grams

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Angelina_Jolie [31]

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