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3 years ago

Can someone help me with 18 to 22plz.

Chemistry

1 answer:

Likurg_2 [28]3 years ago

7 0

Answer:

18 True

19 True

20 False

21 True

Explanation:

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3 years ago

Bromine (Br2) is produced by reacting HBr with O2, with water as a byproduct. The O2 is part of an air (21 mol % O2, 79 mol % N2

Karolina [17]

Answer:

The mole fractions:

$x_{HBr}=\frac{100mol}{318.5}=0.314$

$x_{Br_2}=\frac{78mol}{318.5}=0.245$

$x_{H_2O}=\frac{78mol}{318.5}=0.245$

$x_{O_2}=\frac{62.5mol}{318.5}=0.196$

Explanation:

The reaction described is:

$2 HBr (g) + 1/2 O_2 (g) \longrightarrow Br_2 (g) + H_2O (g)$

The limiting reactant is the HBr (oxygen is in excess).

a) The mass (in moles) balance for this sistem:

$n_{Br_2}=\frac{ 1 mol Br_2}{1 mol HBr} *n_{HBr}*0.78$

(the 0.78 is because of the fractional conversion)

$n_{O_2}=\frac{ 0.5 mol O_2}{1 mol HBr} *n_{HBr}*1.25$

(the 1.25 is because of the oxygen excess)

$n_{H_2O}=\frac{ 1 mol H_2O}{1 mol Br_2} *n_{Br_2}$

There is only one degree of freedom in this sistem, you can either deffine the moles of HBr you have or the moles of Br2 you want to produce. The other variables are all linked by the equations above.

b) Base of calculation 100 mol of HBr:

$nn_{HBr}=100 mol HBr$

$n_{Br_2}=\frac{ 1 mol Br_2}{1 mol HBr} *100mol HBr*0.78$

$n_{Br_2}=78 mol Br2$

$n_{O_2}=\frac{ 0.5 mol O_2}{1 mol HBr} *100 mol HBr*1.25$

$n_{O_2}=62.5 mol O_2$

$n_{H_2O}=n_{Br_2}= 78 mol$

$n_{total}=(78+78+100+62.5)mol= 318.5mol$

The mole fractions:

$x_{HBr}=\frac{100mol}{318.5}=0.314$

$x_{Br_2}=\frac{78mol}{318.5}=0.245$

$x_{H_2O}=\frac{78mol}{318.5}=0.245$

$x_{O_2}=\frac{62.5mol}{318.5}=0.196$

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3 years ago

Urgent!!!! <br> What is number 6 and 12

Tcecarenko [31]

12 is weighting scale

idk 6 tho

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3 years ago

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