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artcher [175]
3 years ago
9

The equilibrium constant for the reaction:

Chemistry
1 answer:
Scrat [10]3 years ago
5 0

Answer:

1. NO and Br₂

2. 77

3. 8.8

Explanation:

Let's consider the following reaction.

2 NO(g) + Br₂(g) ⇄ 2 NOBr(g)

The equilibrium constant for this reaction is:

Kc_{1}=\frac{[NOBr]^{2}}{[NO]^{2}[Br_{2}]} =1.3 \times 10^{-2}

1. At this temperature does the equilibrium favor NO and Br₂, or does it favor NOBr?

Since Kc₁ < 1, the reactants are favored, that is, NO and Br₂.

2. Calculate Kc for 2 NOBr(g) ⇄ 2 NO(g) + Br₂(g)

The equilibrium constant is:

Kc_{2}=\frac{[NO]^{2}[Br_{2}]}{[NOBr]^{2}} =\frac{1}{Kc_{1}} =77

3. Calculate Kc for NOBr(g) ⇄ NO(g) + 1/2 Br₂(g)

The equilibrium constant is:

Kc_{3}=\frac{[NO][Br_{2}]^{1/2} }{[NOBr]} =\sqrt{\frac{[NO]^{2}[Br_{2}]}{[NOBr]^{2}}} =\sqrt{Kc_{2}} =8.8

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3 years ago
The number of mole of a molecule is defined as its mass per molar mass. Using # mol = n, mass = m and molar mass = M and given t
Citrus2011 [14]
Given:
n = 0.0021 mol
M = 44 gram/mol

Required: m or the mass of the molecule

Solution:

M = m/n

Rearranging the expression,

m = M x n
m = 44 g/mol x 0.0021 mol

Evaluating the expression and cancelling units, we obtain,

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3 years ago
A certain liquid X has a normal boiling point of 108.30 °C and a boiling point elevation constant Kb=1.07 °C kg/mol. A solution
Fynjy0 [20]

Answer:

34,6g of (NH₄)₂SO₄

Explanation:

The boiling-point elevation describes the phenomenon in which the boiling point of a liquid increases with the addition of a compound. The formula is:

ΔT = kb×m

Where ΔT is Tsolution - T solvent; kb is ebullioscopic constant and m is molality of ions in solution.

For the problem:

ΔT = 109,7°C-108,3°C = 1,4°C

kb = 1.07 °C kg/mol

Solving:

m = 1,31 mol/kg

As mass of X = 600g = 0,600kg:

1,31mol/kg×0,600kg = 0,785 moles of ions. As (NH₄)₂SO₄ has three ions:

0,785 moles of ions×\frac{1(NH_{4})_{2}SO_{4}}{3Ions} = 0,262 moles of (NH₄)₂SO₄

As molar mass of (NH₄)₂SO₄ is 132,14g/mol:

0,262 moles of (NH₄)₂SO₄×\frac{132,14g}{1mol} = <em>34,6g of (NH₄)₂SO₄</em>

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7 0
3 years ago
PH of 0,035M HCl? How to count if your calculator hasn't got -lg?
IrinaK [193]

<u>Given information:</u>

Concentration of HCl = 0.035 M

<u>To determine:</u>

pH of the solution

<u>Explanation:</u>

Hydrochloric acid, HCl is a strong acid. It will completely dissociate to give H+ and Cl- ions

HCl → H+ + Cl-

Hence the concentration of H+ = Cl- = 0.035M

Now, pH measures the strength of H+ ions in a given solution. It is expressed as:

pH = -log[H+]

pH (HCl) = -log(0.035) = 1.46

Ans: pH of 0.035M HCl is 1.46

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