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Arada [10]
3 years ago
7

The decomposition of a compound at 400⁰C is first order with half-life of 1570 seconds. what fraction of an initial amount of th

e compound remains after 4710secondes?
Chemistry
1 answer:
kirill115 [55]3 years ago
5 0

Answer: After 4710 seconds, 1/8 of the compound will be left

Explanation:

Using the formulae

Nt/No = (1/2)^t/t1/2

Where

N= amount of the compound  present at time t

No= amount of compound present at time t=0

t= time taken for N molecules of the compound to remain = 4710 seconds

t1/2 = half-life of compound  = 1570 seconds

Plugging in the values, we have  

Nt/No = (1/2)^(4710s/1570s)

Nt/No = (1/2)^3

Nt/No= 1/8

Therefore after 4710 seconds, 1/8 molecules of the compound will be left

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Can I get help with this, please!!
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