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3 years ago

How many grams are there in 3.30x10 23 grams of n2i6

1 answer:

8 0

Technically, the answer should be 3.30 * 10^23 grams. But I think you mean either molecules, atoms, moles or grams of Ni2I6 with that number of molecules .

1 mole of Ni2 I6 = 6.02 * 10^23 molecules
x [mole] = 3.30 * 10^23 molecules

1/x = 6.02 * 10^23 / 3.30 * 10^23 Cancel the 10^23 on the right side
1/x = 6.02 / 3.30 Cross multiply
3.30 = 6.02 x Divide by 6.02
3.30 / 6.02 = x
x = 0.548 moles

what to do from here?

1 mole of Ni2I6 is
2 * Ni = 2 * 59 = 118 grams
6 * I = 6 * 131 = 786 grams
Total = 904 grams

Set up a proportion.
1 mole Ni2I6 = 904 grams
0.548 moles = x

1/0.548 = 904/x Cross multiply
x = 0.548 * 904
x = 495.4 grams of Ni2I6 <<<<<< Answer.

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It is called Mass.

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Problem PageQuestion Sulfuric acid is essential to dozens of important industries from steelmaking to plastics and pharmaceutica

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The question is incomplete, here is the complete question:

Sulfuric acid is essential to dozens of important industries from steel making to plastics and pharmaceuticals. More sulfuric acid is made than any other industrial chemical, and world production exceeds 2.0×10¹¹ kg per year.

The first step in the synthesis of sulfuric acid is usually burning solid sulfur to make sulfur dioxide gas. Suppose an engineer studying this reaction introduces 1.8 kg of solid sulfur and 10.0 atm of oxygen gas at 650°C into an evacuated 50.0 L tank. The engineer believes Kp = 0.099 for the reaction at this temperature.

Calculate the mass of solid sulfur he expects to be consumed when the reaction reaches equilibrium. Round your answer to 2 significant digits.

Answer: The mass of solid sulfur that will be consumed is 19. grams

Explanation:

The chemical equation for the formation of sulfur dioxide gas follows:

$S(s)+O_2\rightarrow SO_2(g)$

Initial: 10.0

At eqllm: 10-x x

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{SO_2}}{p_{O_2}}$

We are given:

$K_p=0.099$

Putting values in above expression, we get:

$0.099=\frac{x}{10-x}\\\\x=0.901atm$

Partial pressure of sulfur dioxide = x = 0.901 atm

To calculate the number of moles, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the sulfur dioxide gas = 0.901 atm

V = Volume of the gas = 50.0 L

T = Temperature of the gas = $650^oC=[650+273]K=923K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of sulfur dioxide gas = ?

Putting values in above equation, we get:

$0.901atm\times 50.0L=n\times 0.0821\text{ L. atm}mol^{-1}K^{-1}\times 923K\\\\n=\frac{0.901\times 50.0}{0.0821\times 923}=0.594mol$

By stoichiometry of the reaction:

1 mole of sulfur dioxide gas is produced from 1 mole of sulfur

So, 0.594 moles of sulfur dioxide gas will be produced from = $\frac{1}{1}\times 0.594=0.594mol$ of sulfur

To calculate the number of moles, we use the equation:

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

Moles of sulfur = 0.594 moles

Molar mass of sulfur = 32 g/mol

Putting values in above equation, we get:

$0.594mol=\frac{\text{Mass of sulfur}}{32g/mol}\\\\\text{Mass of sulfur}=(0.594mol\times 32g/mol)=19.008g$

Hence, the mass of solid sulfur that will be consumed is 19. grams

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3 years ago

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Answer:

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Question 1:

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Answer:

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