Condensation releases water
Equation for half-lives:
Nt = No x (1/2)^n
No = initial amount
Nt = final amount after t years
n = number of half lives = t/(single half-life)
t = years
Nt = 3/12 = 0.25
No = 12/12 = 1.00
n = t/(24400)
3/12 = (12/12) x (0.5)^(t/24400)
(0.25) = 1.00 x (0.5)^(t/24400)
0.25/1.00 = 0.5^(t/24400)
ln(0.25) = ln(0.5^(t/24400))
ln(0.25) = (t/24400)*ln(0.5)
ln(0.25)/ln(0.5) = (t/24400)
2 = t/24400
2*24400 = t
t = 48800 yrs
answer is <u>t = 48,800 yrs</u>
Answer:
chemical-Acid rain eating away at the rock
Physical-Dust particles eroding the rock overtime in the wind
Explanation:
Answer: Percent yield represents the ratio between what is experimentally obtained and what is theoretically calculated, multiplied by 100%.
% yield
=
actual yield
theoretical yield
⋅
100
%
So, let's say you want to do an experiment in the lab. You want to measure how much water is produced when 12.0 g of glucose (
C
6
H
12
O
6
) is burned with enough oxygen.
C
6
H
12
O
6
+
6
O
2
→
6
C
O
2
+
6
H
2
O
Since you have a
1
:
6
mole ratio between glucose and water, you can determine how much water you would get by
12.0
g glucose
⋅
1 mole glucose
180.0 g
⋅
6 moles of water
1 mole glucose
⋅
18.0 g
1 mole water
=
7.20
g
This represents your theoretical yield. If the percent yield is 100%, the actual yield will be equal to the theoretical yield. However, after you do the experiment you discover that only 6.50 g of water were produced.
Since less than what was calculated was actually produced, it means that the reaction's percent yield must be smaller than 100%. This is confirmed by
% yield
=
6.50 g
7.20 g
⋅
100
%
=
90.3
%
You can backtrack from here and find out how much glucose reacted
65.0 g of water
⋅
1 mole
18.0 g
⋅
1 mole glucose
6 moles water
⋅
180.0 g
1 mole glucose
=
10.8
g
So not all the glucose reacted, which means that oxygen was not sufficient for the reaction - it acted as a limiting reagent.
Explanation:
