Answer:
406.45mL
Explanation:
The following data were obtained from the question:
V1 = 350mL
P1 = 720mmHg
P2 = 630mmHg
V2 =?
The new volume can be obtain as follows:
P1V1 = P2V2
720 x 350 = 620 x v2
Divide both side by 620
V2 = (720 x 350) /620
V2 = 406.45mL
The new volume of the gas is 406.45mL
Answer:
it is possible to remove 99.99% Cu2 by converting it to Cu(s)
Explanation:
So, from the question/problem above we are given the following ionic or REDOX equations of reactions;
Cu2+ + 2e- <--------------------------------------------------------------> Cu (s) Eo= 0.339 V
Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V
In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:
Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.
Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.
Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.
Thus, ΔG° = - 92640.
This is less than zero[0]. Therefore, it is possible to remove 99.99% Cu2 by converting it to Cu(s) because the reaction is a spontaneous reaction.
The answer is 3.
<span>The relation between number of half-lives (n) and decimal amount remaining (x) can be expressed as:
</span>

We need to calculate n, but we need x to do that. To calculate what p<span>ercentage of a radioactive species would be found as daughter material, we must calculate what amount remained:
1.28 -</span> 1.12 = 0.16
If 1.28 is 100%, how much percent is 0.16:
1.28 : 100% = 0.16 : x
x = 12.5%
Presented as decimal amount:
x = 0.125
Now, let's implement this in the equation:
<span>

</span>
Because of the exponent, we will log both sides of the equation:


<span>

</span>


Therefore, 3 half-lives have passed <span> since the sample originally formed.</span>