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lisov135 [29]
3 years ago
6

What me with this work sheet plllzz

Chemistry
1 answer:
Brut [27]3 years ago
8 0

Answer:

The answer for 17 is maybe

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Which of these is not an example of a molecule?
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Explanation:

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Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir
guajiro [1.7K]

Answer:

y_{O2} =4.3%

Explanation:

The ethanol combustion reaction is:

C_{2}H_{5} OH+3O_{2}→2CO_{2}+3H_{2}O

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}

Dividing the previous equation by x:

1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles

Calculate the number of moles of CO2 and water considering the same:

0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)

0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)

The total number of moles at the reactor output would be:

N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)

So, the oxygen mole fraction would be:

y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3%

6 0
3 years ago
2) Nitrogen gas will react with hydrogen gas to produce ammonia, NH3, How many
Delicious77 [7]

Answer:

12 grams of hydrogen gas

and 56 grams of nitrogen gas

The molar mass of ammonia is 17 g/mol.

68 grams of ammonia corresponds to  

17g/mol

68g

​

=4moles

4 moles of ammonia will be obtained from  

2

4×1

​

=2  moles of nitrogen and  

2

4×3

​

=6  moles of hydrogen.

The molar masses of nitrogen and hydrogen are 28 g/mol and 2 g/mol respectively.

2 moles of nitrogen corresponds to 2×28=56  grams.

6 moles of hydrogen corresponds to 6×2=12  grams.

5 0
2 years ago
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