They are natural resources that are in limited supply or that take a long time to create
<span>an electrolytic cell</span>
The answer is 0.62g.
Solution:
From year 1960 to year 2030, it has been
2030-1960 = 70 years
The half-life of the radioactive element is 28 years, then the sample will go through
70 years * (1 half-life/28 years) = 2.5 half-lives
Starting with a 3.5 gram sample, we will have
3.5*(1/2) after one half-life passes
3.5*(1/2) * (1/2) = 3.5*(1/4) after two half-lives pass
3.5*(1/4) * (1/2) = 3.5*(1/8) after three half-lives pass and so on
Therefore, we can write the remaining amount of the sample after the number n of half-lives have passed as
mass of sample = initial mass of sample/2^n
The mass of the remaining sample for n = 2.5half-lives can be now calculated as
mass of sample = 3.5 grams / 2^2.5 = 0.62 g
Answer:
B. Q > K precipitate will form
Explanation:
The reaction is;
Ba(NO3)2(aq) + Na2CO3(aq) ------> BaCO3(s) + 2NaNO3(aq)
Hence the reaction could form a precipitate of BaCO3.
Number of moles of carbonate ions = 50/1000 * 0.10 M = 5 * 10^-3 moles
Number of moles of Barium ions = 20/1000 * 0.10 M = 2 * 10^-3 moles
Total volume after reaction = 20ml + 50ml = 70 ml or 0.07 L
Molarity Barium ions = 5 * 10^-3 moles/ 0.07 L = 0.07 M
Molarity carbonate ions = 2 * 10^-3 moles/ 0.07 L =0.03 M
Q = [Ba^2+] [CO3^2-] = 0.07 * 0.03 = 2.1 * 10^-3
But K = 2.58 × 10
^−
9
We can clearly see that Q>K therefore precipitate will form
The answer is A.
Hope this helped!!!
