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horsena [70]
3 years ago
13

A sample of ammonium chloride (NH4Cl) has 9.8 moles. How many atoms of hydrogen are in the sample?

Chemistry
1 answer:
blondinia [14]3 years ago
6 0

2.360 × 10^25

Explanation:

The only issue that may arise from asking a question like this is not understanding mole ratios:

Imagine 1 mol of CO2:

You need 1 mol of the end result. In the end result, you have:

1 Carbon

2 Oxygens

When you're trying to get 1 mol of the final compound, you need to have:

1 mol Carbon

2 mol Oxygen

If this doesn't make sense, Khan Academy has insightful videos showing how this works.

So, for NH4Cl, ignore everything but the H. We don't care about them.

You need 4 hydrogens to get 1 NH4Cl, so multiply whatever mole value you're given by 4. That's how many moles of hydrogen there are.

9.8 mol × 4 = 39.2mol H

So the next question is, <em>what</em><em> </em><em>is</em><em> </em><em>a</em><em> </em><em>mole</em><em>?</em><em> </em><em> </em>The answer to this question is the same as asking <em>what</em><em> </em><em>is</em><em> </em><em>a</em><em> </em><em>dozen</em><em>?</em> Well, it's just a number of things. Instead of 12, like a dozen, it's 6.02×10²³. This is called Avogadro's Number.

So say you have two dozen. You'll do 2 × 12, right?

Say you have 39.2moles. Apply the same idea, except with Avogadro's Number instead of 12:

39.2 × 6.02×10²³ = 2.360 × 10^25

So what <em>is</em><em> </em>this number? It's how many <em>things</em><em> </em>are in 39.2moles. We were calculating individual hydrogens, which are atoms, so this number represents how many atoms there are in 39.2moles.

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In a simple distillation setup, what is the sequence of equipment from the bench top to the round bottom flask
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In a simple distillation setup, the sequence of equipment from the bench top to the round bottom flask is:

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<h3>What is Distillation?</h3>

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1 year ago
What mass of Iron (III) acetate will be produced when 1.50 grams of iron reacts in acetic acid?
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Explanation:

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4 0
3 years ago
Calculate the change in entropy when 1.00 kg of water at 100 ∘C is vaporized and converted to steam at 100 ∘C. Assume that the h
andrew11 [14]

Answer : The change in entropy is 6.05\times 10^3J/K

Explanation :

Formula used :

\Delta S=\frac{m\times L_v}{T}

where,

\Delta S = change in entropy = ?

m = mass of water = 1.00 kg

L_v = heat of vaporization of water = 2256\times 10^3J/kg

T = temperature = 100^oC=273+100=373K

Now put all the given values in the above formula, we get:

\Delta S=\frac{(1.00kg)\times (2256\times 10^3J/kg)}{373K}

\Delta S=6048.25J/K=6.05\times 10^3J/K

Therefore, the change in entropy is 6.05\times 10^3J/K

5 0
3 years ago
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