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ololo11 [35]
2 years ago
14

What mass of precipitate will be produced from the reaction of 50.0 mL of 2.50 mol/L sodium hydroxide with an excess of zinc chl

oride solution.
Show your work
Chemistry
1 answer:
amid [387]2 years ago
5 0

2NaOH + ZnCl2 → 2NaCl + Zn(OH)2

Zn(OH)2 is your precipitate.

No. of moles of NaOH = 2.50X50.0÷1000 = 0.125mol

2mol of NaOH reacted to produce 1 mol of Zn(OH)2.

No. of moles of Zn(OH)2 produced when 0.125mol of NaOH reacted

= 0.125 x 1 ÷ 2

= 0.0625mol

Mass of Zn(OH)2 = 0.0625 x [65.4+2(16+1)] = 6.21g

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Svetach [21]

The overall reaction is given by:

Br_{2}(g) + 2 NO(g) \rightarrow  2 NOBr(g)

The fast step reaction is given as:

NO(g) + Br_{2}(g) \rightleftharpoons NOBr_{2}(g) (fast; k_{eq}= \frac{k_{1}}{k_{-1}})

The slow step reaction is given as:

NOBr_{2}(g) + NO(g) \rightarrow  2 NOBr(g) (slow step k_{2})

Now, the expression for the rate of reaction of fast reaction is:

r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]

The expression for the rate of reaction of slow reaction is:

r_{2}=k_{2}[NOBr_{2}] [NO]

Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as [NOBr_{2}] takes place in this reaction.

The expression of rate of formation is:

\frac{d(NOBr)}{dt}=r_{2}

= k_{2}[NOBr_{2}][NO]    (1)

Now, consider that the fast step is always is in equilibrium. Therefore, r_{1}=0

k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]

[NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}]

Substitute the value of [NOBr_{2}] in equation (1), we get:

\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]

=k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]

= \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]

Thus, rate law of formation of NOBr in terms of reactants is given by \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}].









4 0
3 years ago
Two aqueous solutions are prepared: 1.00 m Na2CO3 and 1.00 m LiCl. Which of the following statements is true?
svlad2 [7]

Answer:

The Na_{2}CO_{3} solution has a higher osmotic pressure and higher boiling point than LiCl solution.

Explanation:

As concentrations of two aqueous solutions are same therefore we can write:

                                \Delta P\propto i , \Delta T_{b}\propto i and \pi \propto i

where \Delta P, \Delta T_{b} and \pi are lowering of vapor pressure, elevation in boiling point and osmotic pressure of solution respectively. i is van't hoff factor.

i = total number of ions generated from dissolution of one molecule of a substance (for strong electrolyte).

Here both Na_{2}CO_{3} and LiCl are strong electrolytes.

So, i(Na_{2}CO_{3})=3 and i(LiCl)=2

Hence, lowering of vapor pressure, elevation in boiling point and osmotic pressure will be higher for Na_{2}CO_{3} solution.

Therefore the Na_{2}CO_{3} solution has a higher osmotic pressure and higher boiling point than LiCl solution.

7 0
3 years ago
Dinitrogen tetraoxide, a colorless gas, exists in equilibrium with nitrogen dioxide, a reddish brown gas.
viktelen [127]

Explanation:

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Before proceeding,

A chemical equilibrium can be defined as a condition in the course of a reversible chemical reaction in which no net change in the amounts of reactants and products occurs.

Statement 1.

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Statement 2.

This statement is True in chemical equilibrium; the forward and reverse reactions occur at equal rates.

Statement 3.

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Statement 4.

The concentration of NO2 divided by the concentration of N2O4 is NOT equal to a constant. To obtain a constant value irregardless of the concentrations, the concentration of NO2 must be squared. This comes from the stoichiometry of the reaction

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This statement is false.

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