Equation for Half life :
A = a(0.5)^(t/h)
A is current amount, "a" is initial amount, h is halflife, t is time
5 = 40(0.5)^(t/1.3x10^9)
5/40 = (0.5)^(t/1.3x10^9)
take the log of both sides , power rule
Log(5/40) = (t/1.3x10^9) * Log(0.5)
(1.3x10^9) * Log(5/40) / Log(0.5) = t
3.9x10^9 years = t
And if you think about what a half life is, the time it take for the amount to reduce to half.
40/2 = 20
20/2 = 10
10/2 = 5
It went through 3 half-lifes
3 * 1.3x10^9 = 3.9x10^9 years
Answer:
the answer is longgitudinal
Answer:
42.6 % CHCl₃, 13.83 % C₃H₆O and 43.5 % CH₃COBr
Explanation:
This is a solution with 3 compounds:
96.2 g of CHCl₃
31.2 g of C₃H₆O
98.1 g of CH₃COBr
Total mass of solution is : 96.2g + 31.2 g + 98.1g = 225.5 g
Percent by mass is, the mass of the compounds in 100 g of solution
Let's prepare the rule of three
225.5 g of solution have ____ 96.2 g ___ 31.2 g _____98.1 g
100 g of solution would have ____
(100 . 96.2) / 225.5 = 42.6
(100 . 31.2) / 225.5 = 13.83
(100 . 98.1) / 225.5 = 43.5
Given :
Mass of 
is 571.6 g per liter .
Density of solution , 
.
To Find :
a. Mass percentage
b. Mole fraction
c. Molality
d. molarity of H2SO4 in this solution.
Solution :
Molar mass of , m = 1329 g/mol .
a ) Mass of contain in 1 liter is 1329 g .
b ) Moles of = 
.
Moles of 
= 
.
Mole fraction 
.
c ) Molarity of 
.
Hence , this is the required solution .
