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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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How many molecules are in 7.62 L of CH4, at 87.5°C and 722 torr
pickupchik [31]

Answer: There are 1.469 \times 10^{23} molecules present in 7.62 L of CH_4 at 87.5^{o}C and 722 torr.

Explanation:

Given : Volume = 7.62 L

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Converting torr into atm as follows.

722 torr = 722 torr \times \frac{0.00131579 atm}{1 torr}\\= 0.95 atm

Therefore, using the ideal gas equation the number of moles are calculated as follows.

PV = nRT

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V = volume

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R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

PV = nRT\\0.95 atm \times 7.62 L = n \times 0.0821 L atm/mol K \times 360.5 K\\n = \frac{0.95 atm \times 7.62 L}{0.0821 L atm/mol K \times 360.5 K}\\= \frac{7.239}{29.59705}\\= 0.244 mol

According to the mole concept, 1 mole of every substance contains 6.022 \times 10^{23} atoms. Hence, number of atoms or molecules present in 0.244 mol are calculated as follows.

0.244 mol \times 6.022 \times 10^{23}\\= 1.469 \times 10^{23}

Thus, we can conclude that there are 1.469 \times 10^{23} molecules present in 7.62 L of CH_4 at 87.5^{o}C and 722 torr.

5 0
3 years ago
What mass of Hg will occupy a volume of 75.0 mL?
k0ka [10]
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<span>0.00334 moles Hg ----- ( mass Hg )
</span>
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mass Hg = 0.6699 / 1

= 0.6699 g of Hg




7 0
3 years ago
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