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dusya [7]
2 years ago
6

The rate law for the decomposition of phosphine () is It takes 135. s for 1.00 M to decrease to 0.250 M. How much time is requir

ed for 3.00 M to decrease to a concentration of 0.450 M
Chemistry
1 answer:
vova2212 [387]2 years ago
8 0

Answer:

189.71 secs

Explanation:

We know that decomposition is a first order reaction;

So;

ln[A] = ln[A]o - kt

But;

[A]o = 1.00 M

[A] = 0.250 M

t =135 s

Hence;

ln[A] -  ln[A]o = kt

k = ln[A] -  ln[A]o/t

k = ln(1) - ln(0.250)/135

k =0 - (-1.386)/135

k = 1.386/135

k= 0.01

So time taken now will be;

ln[A] -  ln[A]o = kt

t = ln[A] -  ln[A]o/k

t = ln (3) - ln(0.450)/0.01

t = 1.0986 - (-0.7985)/0.01

t = 1.0986 + 0.7985/0.01

t = 189.71 secs

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A) What is the maximum number of grams of nickel bromide that can be produced from the reaction of 67.8 g of nickel with 37.3 g
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The answer to your question is a) 51.07 g of NiBr₂   b) Nickel, 54 g

Explanation:

Data

mass of NiBr₂ = ?

mass if Ni = 67.8 g

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Balanced chemical reaction

                Ni  +  Br₂   ⇒   NiBr₂

Process

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The limiting reactant is Bromine because the experimental yield was lower than the theoretical yield.

3.- Calculate the mass of NiBr₂

                    159.8 g of Br₂ --------------- 218.8 g of NiBr₂

                      37.3 g of Br₂ --------------  x

                          x = (37.3 x 218.8) / 159.8

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                  x               ----------------  37.3 g of Br₂

                            x = (37.3 x 59)/159.8

                            x = 2200.7/159.8

                            x = 13.77 g of Ni

Excess Ni = 67.8 - 13.77

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