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sergiy2304 [10]
3 years ago
13

If a mineral has a daughter to parent ration of 3:1 (three daughter atoms for every one parent atom), how many half-lives have p

assed since the mineral formed?
Chemistry
1 answer:
kirill [66]3 years ago
4 0
The answer to this question would be: 2 half-lives

In this question, the mineral has a daughter to parent ratio of 3:1. That means, the amount of parent mass should be are 1/(3+1)= 1/4 of the total mass. For every half live, the parent mass would become 1/2. Then for 1/4, the number of half-lives should be:
1/2^x= 1/4
1/2^x= 1/2^2
x=2
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Answer:

The final balanced equation is :

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Explanation:

SO_4^{2-}(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+Sn^{4+}(aq)

Balancing in acidic medium:

First we will determine the oxidation and reduction reaction from the givne reaction :

Oxidation:

Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)

Balance the charge by adding 2 electrons on product side:

Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)+2e^-....[1]

Reduction :

SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)

Balance O by adding water on required side:

SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)

Now, balance H by adding H^+ on the required side:

SO_4^{2-}(aq)+4H^+(aq)\rightarrow H_2SO_3(aq)+H_2O(l)

At last balance the charge by adding electrons on the side where positive charge is more:

SO_4^{2-}(aq)+4H^+(aq)+2e^-\rightarrow H_2SO_3(aq)+H_2O(l)..[2]

Adding [1] and [2]:

SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)

The final balanced equation is :

SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)

4 0
3 years ago
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Answer:

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Rock cycle(Attachment-1)

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Explanation:

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