It would contain 0.105M moles of NaOH in each liter of solution
<u>Answer: </u>
A sample initially contained 150 mg of radon-222. After 11.4 days only 18.75mg of the radon-222 in the sample remained where 3 half-lives have passed
<u>Explanation:</u>
Given, the initial value of the sample,
= 150mg
Final value of the sample or the quantity left, A = 18.75mg
Time = 11.4 days
The amount left after first half life will be ½.
The number of half-life is calculated by the formula

where N is the no. of half life
Substituting the values,


On equating, we get, N = 3
Therefore, 3 half-lives have passed.
Answer: cause oxegeon is harmful then helium
Explanation:
6.4 x 10^-7 = [CO]^2[O2]/ [CO2}^2 = ( 2.0 x 10^-3)^2 ( 1.0 x 10^-3)/ [CO2]^2 =
<span>=4.0 x 10^-9 / [CO2]^2 </span>
<span>[CO]= sq.rt ( 4.0 x 10^-9)/ 6.4 x 10^-7=7.9 x 10^-2 M </span>
<span>2.6 x 10^-3 = [I]^2 / [I2] = [I]^2 / 0.95 </span>
<span>[I]= sq.rt ( 2.6 x 10^-3 x 0.95)=5.0 x 10^-2 M </span>
<span>Ksp = [Ba2+][CO32-] = ( 1.1 x 10^-4)^2=1.2 x 10^-8</span>