Answer:
The percentage of the block contains 15% after 60 years.
Explanation:
Step 1: Formula for half-life time
To calculate the half-life time we will use the following formula:
At = A0 * 1/2 ^(x/t)
With At = the quantity after a time t
A0 = The quantity at time t = 0 (start)
x = time in this case = 60 years
t= half-life time = 22 years
Step 2: Calculate the percentage after 60 years
In this case: after 60 years the percentage will be
A0= 10 * 1/2 ^(60/22)
A0 = 1.5
A0 / At = 1.5 /10 = 0.15
0.15 *100% = 15 %
The percentage of the block contains 15% after 60 years.
Answer:
d. Radio wave
Explanation:
There are three types of mechanical waves, transverse waves, longitudinal waves, and surface waves
This is probably wrong but i think the answer is 1,979,381.44
After 1 half life, 1/2
After 2 half lives, 1/4
After 3 half lives, 1/8
Answer:
<em> 14, 508J/K</em>
ΔHrxn =q/n
where q = heat absorbed and n = moles
Explanation:
<em>m = mass of substance (g) = 0.1184g</em>
1 mole of Mg - 24g
<em>n</em> moles - 0.1184g
<em>n = 0.0049 moles.</em>
Also, q = m × c × ΔT
<em> Heat Capacity, C of MgCl2 = 71.09 J/(mol K)</em>
<em>∴ specific heat c of MgCL2 = 71.09/0.0049 (from the formula c = C/n)</em>
<em>= 14, 508 J/K/kg</em>
ΔT= (final - initial) temp = 38.3 - 27.2
= 11.1 °C.
mass of MgCl2 = 95.211 × 0.1184 = 11.27
⇒ q = 11.27g × 11.1 °C × <em>14, 508 j/K/kg </em>
<em>= 1,7117.7472 J °C-1 g-1</em>
<em />
<em>∴ ΔHrxn = q/n</em>
<em>=1,7117.7472 ÷ 0.1184 </em>
<em>= 14, 508J/K</em>