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Rina8888 [55]
3 years ago
11

Which phase of matter has a fixed volume but variable shape

Chemistry
1 answer:
anastassius [24]3 years ago
5 0
The answer is the liquid phase
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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

4 0
3 years ago
I'm deaf and help 6th grader
Katena32 [7]

Answer:

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Explanation: next time add description please so that i can see

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