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3 years ago

How many moles of sulfur atoms are in 9.9 moles of Ag2S

Chemistry

1 answer:

GarryVolchara [31]3 years ago

6 0

Answer:

9.9 moles of Ag₂S contain 9.9 moles of sulfur.

Explanation:

Given data:

Number of moles of Ag₂S = 9.9 mol

Number of moles of Sulfur = ?

Solution:

One mole of Ag₂S having 1 mole of sulfur.

In 9.9 moles:

9.9 moles × 1 = 9.9 moles

So 9.9 moles of Ag₂S contain 9.9 moles of sulfur.

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39. Which of the following substances are insoluble in water? Select all that apply.

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3 years ago

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What Happens During Fermentation?

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Answer:

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2 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

3 years ago

Hydrogen peroxide can decompose to water and oxygen by the following reaction

mestny [16]

To begin calculating, there is one thing you need to remember : $1 mole of H2O2=34.0148 g$
Then we have $5.00g of H2O2=5/34.0148=0.146995$
As you know decomposition of 2moles now has prodused 196kj
So, q is made due $0.146995 moles of H2O2$ $=(-196/2)*0.146995=-14.40551Kj$
I'm sure it will help.

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3 years ago