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3 years ago

5

How many moles of water are present in 55.1 g of H2O

1 answer:

Rashid [163]3 years ago

7 0

Answer:

3.0585147719047385 is the answer

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Elecstrostatic force

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Why does dark chocolate melt faster in the sun?<br><br> Claim<br><br> Evidence <br><br> Reasoning

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3 years ago

Which statements describe a mushroom’s niche? Check all that apply.

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3 years ago

Read 2 more answers

An experimenter found that for the hydrolysis of lactose, ΔH° = +0.44 kJ mol–1 and ΔS° = +0.031 kJ mol–1 K–1. ΔG°for this reacti

Ksivusya [100]

Answer:

ΔG° = -8.8 kJ/mol

Explanation:

The standard Gibbs free energy of reaction (ΔG°) can be calculated using the following expression.

ΔG° = ΔH° - T.ΔS°

where,

ΔH°: standard enthalpy of reaction

T: absolute temperature

ΔS°: standard entropy of reaction

At 298 K (the temperature that is usually used), ΔG° for the hydrolysis of lactose is:

ΔG° = ΔH° - T.ΔS°

ΔG° = 0.44 kJ/mol - 298 K × 0.031 kJ/mol.K

ΔG° = -8.8 kJ/mol

7 0

3 years ago

At 920 K, Kp = 0.40 for the following reaction. 2 SO2(g) + O2(g) equilibrium reaction arrow 2 SO3(g) Calculate the equilibrium p

Alex777 [14]

<u>Answer:</u> The equilibrium partial pressure of sulfur dioxide, oxygen and sulfur trioxide is 0.366 atm, 0.443 atm and 0.154 atm respectively.

<u>Explanation:</u>

We are given:

Initial partial pressure of sulfur dioxide = 0.52 atm

Initial partial pressure of oxygen = 0.52 atm

Initial partial pressure of sulfur trioxide = 0 atm

For the given chemical reaction:

$2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$

<u>Initial:</u> 0.52 0.52

<u>At eqllm:</u> 0.52-2x 0.52-x 2x

The expression of $K_p$ for above equation follows:

$K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times p_{O_2}}$

We are given:

$K_p=0.40\\\\p_{SO_3}=2x\\\\p_{SO_2}=0.52-2x\\\\p_{O_2}=0.52-x$

Putting values in above equation, we get:

$0.40=\frac{(2x)^2}{(0.52-2x)^2\times (0.52-x)}\\\\x=-1.13,-0.402,0.077$

Neglecting the negative values because partial pressure cannot be negative.

So, x = 0.077

Equilibrium partial pressure of sulfur dioxide = $(0.52-2x)=(0.52-(2\times 0.077))=0.366atm$

Equilibrium partial pressure of oxygen = $(0.52-x)=(0.52-0.077)=0.443atm$

Equilibrium partial pressure of sulfur trioxide = $2x=(2\times 0.077)=0.154atm$

Hence, the equilibrium partial pressure of sulfur dioxide, oxygen and sulfur trioxide is 0.366 atm, 0.443 atm and 0.154 atm respectively.

4 0

3 years ago