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Chemistry

1 answer:

Ivanshal [37]4 years ago

6 0

Answer:

The answer D. C2H5COOH

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Please help due today!

Oksana_A [137]

Answer:

it’s 2

Explanation:

yesssss

5 0

3 years ago

Balance the following equation: * 1, 1, 1 1, 3, 1 2, 3, 2 1, 3, 2

alexira [117]

1,3,2

The coefficient multiplied by the subscript will give you how many atoms of the element there are.

So, if we plug in 1,3,2
There will be 2 Nitrogen and 6 Hydrogen in the reactants side of the equation. To balance this on the other side, we must put a 2 in order to get 2 Nitrogen and 6 Hydrogen again.

6 0

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:

$nn_{HBr}=100 mol HBr$