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strojnjashka [21]
3 years ago
15

What is the empirical formula for a substance that contains 3.730% hydrogen, 44.44% carbon, and 51.83% nitrogen by mass?\?

Chemistry
1 answer:
Vlada [557]3 years ago
3 0

Given data:

Hydrogen (H) = 3.730 % by mass

Carbon (C) = 44.44%

Nitrogen (N) = 51.83 %

This means that if  the sample weighs 100 g then:

Mass of H = 3.730 g

Mass of C = 44.44 g

Mass of N = 51.83 g

Now, calculate the # moles of each element:

# moles of H = 3.730 g/ 1 g.mole-1 = 3.730 moles

# moles of C = 44.44/12 = 3.703 moles

# moles of N = 51.83/14 = 3.702 moles

Divide by the lowest # moles:

H = 3.730/3.702  = 1

C = 3.703/3.702 = 1

N = 3.702/3.702 = 1

Empirical Formula = HCN

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Answer:

34 g

Explanation:

Let's consider the following balanced equation.

N₂ + 3 H₂ → 2 NH₃

The theoretical mass ratio of N₂ to H₂ is 28g N₂ : 6g H₂ = 4.6g N₂ : 1g H₂.

The experimental mass ratio of N₂ to H₂ is 100g N₂ : 6g H₂ = 16.6g N₂ : 1g H₂.

As we can see, hydrogen is the limiting reactant.

According to the task, we 6 g of H₂ react completely, 34 g of ammonia are produced.

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How many moles are in 400 ml of hydrogen
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Answer:

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Explanation:

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The density of gold is 19.3 g/cm3. What is the volume of a 575 gram bar of pure gold?
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Answer:

The  answer to your question is: Volume = 29.79 cm³

Explanation:

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