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3 years ago

It’s a test and only need 2 more problems!!!! Thank you and Appreciate it!!!

Chemistry

1 answer:

Mkey [24]3 years ago

3 0

Answer:

the molar mass of propane (c3H8) is

12*3+1*8

=36+8=44

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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

What volume of hydrogen ( in L ) is produced from the complete reaction of 56.49 g of magnesium metal at STP

alexandr1967 [171]

Answer:

<em>Mg = 24.30 g/mol) Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) Hint: 1 mole of gas at STP occupies 22.4 L</em>

6 0

2 years ago

Could someone help me with this?

Varvara68 [4.7K]

Solving part-1 only

KMnO_4

Transition metal is Manganese (Mn)

Actually it's the oxidation number of Mn

Let's find how?

$\\ \tt\Rrightarrow x+1+4(-2)=0$

$\\ \tt\Rrightarrow x+1-8=0$

$\\ \tt\Rrightarrow x-7=0$

$\\ \tt\Rrightarrow x=+7$

x is the oxidation number

Purple as per the color of potassium permanganate

$\boxed{\begin{array}{c|c|c}\boxed{\bf Tube} &\boxed{\bf Charge} &\boxed{\bf No\:of\; electrons\: loss}\\ \sf 2 &\sf +6 &\sf 6e^-\\ \sf 3& \sf +2 &\sf 2e- \\ \sf 4 &\sf 4 &\sf 4e^-\end{array}}$

7 0

2 years ago

A student is adding DI water to a volumetric flask to make a 50% solution. Unfortunately, he was not paying attention and filled

dalvyx [7]

Answer:

His results will be skewed because there was more water than stock solution. Which would cause the percentage solution to be less than 50% therefore the density would be less than the actual value.

Explanation:

The solution will have percentage less than that of 50%. Therefore the density would be less than the actual value.

Suppose there should be 50 mL of the solution, and he added 60 mL. So 10 mL of the solution is added more.

Suppose the mass of the solute is m.

Originally, the density is = $\frac{m}{50}$ $\left(\frac{\text{mass}}{\text{volume}}\right)$

Now after adding extra 10 mL , the density becomes $\frac{m}{60}$ .

Therefore, $\frac{m}{50}>\frac{m}{60}$

So the density decreases when we add more solution.

4 0

3 years ago

Why does the same side of the moon always face earth? A. The rotational period of earth is the same as that of the moon B. The r

mixer [17]

Answer: A. the rotational period of the earth is the same as that of the moon

3 0

3 years ago

Read 2 more answers