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Burka [1]
3 years ago
14

An unknown compound contains only carbon, hydrogen, and oxygen (). Combustion of 7.50 of this compound produced 11.0 of carbon d

ioxide and 4.50 of water.
1) How many moles of carbon, C, were in the original sample?
An unknown compound contains only carbon, hydrogen, and oxygen (). Combustion of 7.50 of this compound produced 11.0 of carbon dioxide and 4.50 of water.

1) How many moles of carbon, C, were in the original sample?
Chemistry
1 answer:
Sergeu [11.5K]3 years ago
5 0
<span>Firstly, we know that M= m/n,  the main formula which shows the relationship between m, n, and M. The nknown compound contains only carbon, hydrogen, and oxygen, so we can get n(C)=m/M,  from M(C)= m(C)/n (C),  besides the stoechiometric equality, we have </span>

n( C)= m(C)/M(C ) = m(CO2)/ M(CO2)=11/44, because m(CO2)=11.0,  M(CO2)=44.01

so n(C )= 0.24moles,




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Answer:

Option B. 4NH3(g) + 5O2(g) —> 4NO(g) + 6H2O(g)

Explanation:

The reaction between ammonia, NH3 and oxygen, O2, will produce nitric oxide and water as shown below:

NH3(g) + O2(g) —> NO(g) + H2O(g)

Now, let us balance the equation.

This is illustrated below:

NH3(g) + O2(g) —> NO(g) + H2O(g)

There are 3 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of NH3 and 6 in front of H2O as shown below:

4NH3(g) + O2(g) —> NO(g) + 6H2O(g)

There are 4 atoms of N on the left side and 1 atom on the right side. It can be balance by putting 4 in front of NO as shown below:

4NH3(g) + O2(g) —> 4NO(g) + 6H2O(g)

Now, the are 2 atoms of O on the left side and a total of 10 atoms on the right side. It can be balance by putting 5 in front of O2 as shown below:

4NH3(g) + 5O2(g) —> 4NO(g) + 6H2O(g)

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Answer:

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Explanation:

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Answer:

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Explanation:

Step 1: Data given

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Volume = 2.0 L

Step 2: The balanced equation

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Step 3: Calculate molarity

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Kc = 1/0,.000000000125

Kc = 8.0 * 10^9

Step 5: Calculate Kp

Kp = Kc*(R*T)^Δn

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⇒with R = 0.08206 L*atm /mol*K

⇒with T = 298 K

⇒with Δn = -3

Kp = 8.10^9 *(0.08206 * 298)^-3

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