
60g after 3 hours, 30g after 6 hours and 15g after 9 hours
Explanation:
Weight of the radioactive sample = 120g
half life time period = 3 hours
(a) The weight of sample after 3 hours

The fraction of sample left

Mass of the sample left

<u>6</u><u>0</u><u>g</u><u> </u><u>of</u><u> </u><u>sample</u><u> </u><u>is</u><u> </u><u>left</u><u> </u><u>after</u><u> </u><u>3</u><u> </u><u>hours</u>
(b) The weight of sample after 6 hours

The fraction of the sample left

Mass of the sample left

<u>3</u><u>0</u><u>g</u><u> </u><u>of</u><u> </u><u>sample</u><u> </u><u>is</u><u> </u><u>left</u><u> </u><u>after</u><u> </u><u>6</u><u> </u><u>hours</u>
(c) The weight of sample after 9 hours

The fraction of sample left

Mass of sample left

<u>1</u><u>5</u><u>g</u><u> </u><u>of</u><u> </u><u>sample</u><u> </u><u>is</u><u> </u><u>left</u><u> </u><u>after</u><u> </u><u>9</u><u> </u><u>hours</u><u>.</u>
Try G I think I’m correct
Ozone 03 have 2 sigma bond and 1 pie bond.
just remember this and you can do any problem for pie and sigma with this.
single bond = 1 sigma
double bond = 1 sigma and 1 pie
triple bond = 1 signa and 2 pie
That is false. mass is converted into energy
Answer:
116 years
Explanation:
To solve this, we will use the half life equation;
A(t) = A_o(½)^(t/t_½)
Where;
A(t) is the amount of strontium left after t years;
A_o is the initial quantity of strontium that will undergo decay;
t_½ is the half-life of strontium
t is the time it will take to decay
We are given;
A(t) = 7.5 g
A_o = 120 g
From online values, half life of strontium-90 is 29 years. Thus, t_½ = 29
Thus;
7.5 = 120 × ½^(t/29)
Divide both sides by 120 to get;
7.5/120 = ½^(t/29)
0.0625 = ½^(t/29)
In 0.0625 = (t/29) In ½
-2.772589 = (t/29) × (-0.693147)
(t/29) = -2.772589/(-0.693147)
t/29 = 4
t = 29 × 4
t = 116 years