__MgF2 + __Li2CO3 + __ 2LiF
The amount of substance present in a certain object with a given half-life in terms of h can be expressed through the equation,
A(t) = (A(o))(0.5)^(t/h)
where A(t) is the amount of substance after t years and A(o) is the original amount. In this item we are given that A(t)/A(o) is equal to 0.89. Substituting the known values,
0.89 = (0.5)(t / 5730 years)
The value of t from the equation is 963.34 years.
<em>Answer: 963 years</em>
Answer:
New volume of gas = 95.93 ml (Approx)
Explanation:
Given:
Old volume of gas = 86 ml
Old temperature = 30°C = 30 + 273 = 303 K
New temperature = 65°C = 65 + 273 = 338 K
Find:
New volume of gas
Computation:
V1T2 = V2T1
(86)(338) = (V2)(303)
New volume of gas = 95.93 ml (Approx)
Answer : The concentration of A after 80 min is, 0.100 M
Explanation :
Half-life = 20 min
First we have to calculate the rate constant, we use the formula :
Expression for rate law for first order kinetics is given by:
where,
k = rate constant = 
t = time passed by the sample = 80 min
a = initial amount of the reactant = 1.6 M
a - x = amount left after decay process = ?
Now put all the given values in above equation, we get
Therefore, the concentration of A after 80 min is, 0.100 M
