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3 years ago

Determine the average rate of decomposition of h3po4 between 10.0 and 40.0 s. express your answer with the appropriate units.

1 answer:

8 0

Assume that we have this reaction 2H3PO4⇒ P2O5 + 3H2O
and when the rate of reaction is the speed at which the reactants can be changed and are converted into products So, according to our balanced reaction:

The rate of reaction = rate of formation of P2O5
and the rate of this reaction also = 0.5 rates of decomposition of H3PO4
∴ the rate of decomposition of H3PO4 = 2 rates of formation of P2O5
So when the rate of formation P2O5 = 2 x 10^-4 M/s
So the rate of decomposition of H3PO4 = 2 x 2 x 10^-4
= 4 x 10^-4 M/s

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3 years ago

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3 years ago

How many half-lives will pass by the time 1.56% of I-131 is present? B. Approximately how many days does that equal? *

serg [7]

Answer: Hmmmmm that's crazy....

There are a couple of equations one could use for this type of problem, but I find the following to be the easiest to use and to understand.

Fraction remaining (FR) = 0.5n

n = number of half lives that have elapsed

In this problem, we need to find n and are given the FR, which is 1.56% or 0.0156 (as a fraction).

0.0156 = 0.5n

log 0.0156 = n log 0.5

-1.81 = -0.301 n

n = 6.0 half lives have elapsed

Explanation:

Just wanted to help. Hopefully it's correct wouldn't want to waster your time ;)

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2 years ago

In each case, calculate the appropriate ratio to show that the information given is consistent with the law of multiple proporti

Alchen [17]

Answer:

(a) 3:2; (b) 2:1

Explanation:

The Law of Multiple Proportions states that when two elements A and B combine to form two or more compounds, the masses of B that combine with a given mass of A are in the ratios of small whole numbers.

That is, if one compound has a ratio r₁ and the other has a ratio r₂, the ratio of the ratios r is in small whole numbers.

(a) Ammonia and hydrazine.

In ammonia, the mass ratio of H:N is r₁ = 0.2158/1

In hydrazine, the mass ratio of H:N is r₂ = 0.1439/1

The ratio of the ratios is:

$r = \dfrac{r_{1}}{r_{2}} = \dfrac{ 0.2158}{0.1439} = \dfrac{1.500 }{1} = \dfrac{2.999}{2} \approx \mathbf{\dfrac{3}{2}}\\\\\text{The relative amounts of H in the two compounds are in the ratio }\boxed{\mathbf{\dfrac{3}{2}}}$

(b) Nitrogen oxides

In nitrogen monoxide, the mass ratio of O:N is r₁ = 1.142/1

In dinitrogen monoxide, the mass ratio of O:N is r₂ = 0.571/1

The ratio of the ratios is:

$r = \dfrac{r_{1}}{r_{2}} = \dfrac{ 1.142}{0.571} = \dfrac{2.000 }{1} \approx \mathbf{\dfrac{2}{1}}\\\\\text{The relative amounts of O in the two compounds are in the ratio }\boxed{\mathbf{\dfrac{2}{1}}}$

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3 years ago