The half life of carbon-14 is 5700 years. (Half life is the time taken by a radioactive isotope to decay by half of its original mass).
Let A₀ be the initial amount of carbon-14 that is found in living matter (t=0 years), to determine when there was 44.5% of A₀ left.
44.5 = 100 × (1/2)^n, where n is the number of half lives
0.5^n = 0.445
n = log 0.445/log 0.5
n = 1.168
But 1 half life is 5700 years
Therefore, the number of years will be 5700 × 1.168 = 6658.299725 years
≈ 6658.30 years
Answer:
hr
Explanation:
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Answer is: 1.29 grams <span>of solid formed.
</span>Chemical reaction: 2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq).
n(AgNO₃) = c(AgNO₃) · V(AgNO₃).
n(AgNO₃) = 0.220 M · 0.0351 L.
n(AgNO₃) = 0.0078 mol; limiting reactant.
n(K₂CrO₄) = 0.420 M · 0.052 L.
n(K₂CrO₄) = 0.022 mol.
From chemical reaction: n(AgNO₃) : n(Ag₂CrO₄) = 2 : 1.
n(Ag₂CrO₄) = 0.0078 mol ÷ 2.
n(Ag₂CrO₄) = 0.0039 mol.
m(Ag₂CrO₄) = 0.0039 mol · 331.73 g/mol.
m(Ag₂CrO₄) = 1.29 g.
This is an oxidation reduction reaction
oxidation happens when the species gives out electrons.Oxidation state of the element Cu increases from 0 to +2
Cu ---> Cu²⁺ +2e --1)
Reduction happens when the species gains electrons.
Oxidation state of N reduces from +5 to +2
4H⁺ + NO₃⁻ + 3e ---> NO + 2H₂O --2)
To balance the reactions, number of electrons need to be balanced.
Multiply 1st reaction by 3
Multiply 2nd reaction by 2
3Cu ---> 3Cu²⁺ + 6e
8H⁺ + 2NO₃⁻ + 6e ---> 2NO + 4H₂O
add the 2 equations
3Cu + 8H⁺ + 2NO₃⁻ --> 3Cu²⁺ + 2NO + 4H₂O
add 6NO₃⁻ ions to each side
3Cu + 8HNO₃ --> 3Cu(NO₃)₂ + 2NO + 4H₂O
the balanced redox reaction equation is as follows;
3Cu(s) + 8HNO₃(aq) --> 3Cu(NO₃)₂ (aq)+ 2NO(g) + 4H₂O(l)
