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3 years ago

Hydrogen sulfide burns form sulfur dioxide:

Chemistry

1 answer:

Helga [31]3 years ago

7 0

Answer: 404.04 kJ.

Explanation:

To calculate the moles, we use the equation:

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

moles of $H_2S$

$\text{Number of moles}=\frac{26.7g}{34.1g/mol}=0.78moles$

$2H_2S(g)+3O_2(g)\rightarrow 2SO_2(g)+2H_2O(g)$ $\Delta H=-1036kJ$

According to stoichiometry :

2 moles of $H_2S$ on burning produces = 1036 kJ

Thus 0.78 moles of $H_2S$ on burning produces = $\frac{1036kJ}{2}\times 0.78=404.04$

Thus the enthalpy change when burning 26.7 g of hydrogen sulfide is 404.04 kJ.

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Then at equilibrium cations and anions concentration is considered same hence:

$\left[\mathbf{A} \mathbf{g}^{+}\right]=[\mathbf{C} \mathbf{I}]=\text { molarity of ions }$

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Answer:

$3.4\times 10^3 eggs$ are produces are in a month.

Explanation:

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3 years ago

Consider the following chemical reaction:

laiz [17]

Answer:

B. 1.65 L

Explanation:

Step 1: Write the balanced equation

2 SO₂(g) + O₂(g) ⇒ 2 SO₃(g)

Step 2: Calculate the moles of SO₂

The pressure of the gas is 1.20 atm and the temperature 25 °C (298 K). We can calculate the moles using the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

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Step 3: Calculate the moles of SO₃ produced

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Step 4: Calculate the volume occupied by 0.0736 moles of SO₃ at STP

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2 years ago