The time taken by Carbon-14 to decay radioactively from 120g to 112.5g is 22,920 years.
<h3>How do we calculate the total time of decay?</h3>
Time required for the whole radioactive decay of any substance will be calculated by using the below link:
T = (n)(t), where
- t = half life time = 5730 years
- n = number of half life required for the decay
Initial mass of Carbon-14 = 120g
Final mass of Carbon-14 = 112.5g
Left mass = 120 - 112 = 7.5g
Number of required half life for this will be:
- 1: 120 → 60
- 2: 60 → 30
- 3: 30 → 15
- 4: 15 → 7.5
4 half lives are required, now on putting values we get
T = (4)(5730) = 22,920 years
Hence required time for the decay is 22,920 years.
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To fully understand the problem, we use the ICE table to identify the concentration of the species. We calculate as follows:
Ka = 2.0 x 10^-9 = [H+][OBr-] / [HOBr]
HOBr = 0.50 M
KOBr = 0.30 M = OBr-
<span> HOBr + H2O <-> H+ + OBr- </span>
<span>I 0.50 - 0 0.30 </span>
<span>C -x x x
</span>---------------------------------------------
<span>E(0.50-x) x (0.30+x) </span>
<span>Assuming that the value of x is small as compared to 0.30 and 0.50 </span>
<span>Ka = 2.0 x 10^-9 = x (0.30) / 0.50) </span>
<span>x = 3.33 x 10^-9 = H+</span>
pH = 8.48
Answer:
Correct choice are C and D (they are both, the same).
Explanation:
Cathode is the positive(+) electrode where a reduction occurs.
Reduction is the chemical reaction where the oxidation state is reduced.
2Ag(s) + 1/2 O2(g) + 2H+(aq) → 2Ag+(aq) + H2O (l)
A. 2H2O (l) → O2 (g) + 4H+ (aq) + 4e-
B. 2Ag (s) → 2Ag+ (aq) + 2e-
C. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)
D. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)
C or D, are ok. They are the same equation.
Oxygen from ground state reduce the oxidation state from 0 to -2
It 1
cause yea it is im doing it in class and the answers is 1
