The half life for C14 is 5730 years.
We assume that Carbon 14/ Carbon 12 ratio was steady for living organisms over time, the problem is actually telling us that
= 0.0725 =
ˣ
Take the natural logarithm and In on both sides.
ln(0.725) = ln
ˣ
= - 0.3216 = xln (
= -0.6931x.
So x = (-.3216) / (-0.6931) = 0.464
or
t/t₁/₂ = 0.464
So t = 0.464 x t₁/₂ = 0.464 * 5730 yrs = 2660 years.
1. 12 L = 12 dm³
2. 3.18 g
<h3>Further explanation</h3>
Given
1. Reaction
K₂CO₃+2HNO₃⇒ 2KNO₃+H₂O+CO₂
69 g K₂CO₃
2. 0.03 mol/L Na₂CO₃
Required
1. volume of CO₂
2. mass Na₂CO₃
Solution
1. mol K₂CO₃(MW=138 g/mol) :
= 69 : 138
= 0.5
mol ratio of K₂CO₃ : CO₂ = 1 : 1, so mol CO₂ = 0.5
Assume at RTP(25 C, 1 atm) 1 mol gas = 24 L, so volume CO₂ :
= 0.5 x 24 L
= 12 L
2. M Na₂CO₃ = 0.03 M
Volume = 1 L
mol Na₂CO₃ :
= M x V
= 0.03 x 1
= 0.03 moles
Mass Na₂CO₃(MW=106 g/mol) :
= mol x MW
= 0.03 x 106
= 3.18 g
Answer:
A. Up
B. Out
C. Out
D. To equilibrum
Explanation:
a. The reaction in an exothermic reaction so this means heat is given off. If the cylinder is thin enough heat will transfer to the water bath
b. Since the products will create heat which will increase pressure, the piston in an attempt to maintaining a constant pressure will move up to accommodate building pressure.
c. Heat will flow out of the gaseous mixture as this reaction creates heat as a product as well
d. Heat will flow out in the capacity to create an equilibrium with the water bath that it is in.
Answer : The half-life of the compound is, 145 years.
Explanation :
First we have to calculate the rate constant.
Expression for rate law for first order kinetics is given by:
where,
k = rate constant = ?
t = time passed by the sample = 60.0 min
a = let initial amount of the reactant = 100 g
a - x = amount left after decay process = 100 - 25 = 75 g
Now put all the given values in above equation, we get
Now we have to calculate the half-life of the compound.
Therefore, the half-life of the compound is, 145 years.
