The answer is B). A wave loses energy as it moves through a medium.
I don’t know I just need to ask a question
Answer is: 12,6% (1/8) <span>percentage of the sample will remain.
</span>c₀ - initial amount of C-14.<span>
c - amount of C-14 remaining
at time.
t = 5700</span> y.<span>
First calculate the radioactive decay rate constant λ:
λ = 0,693 ÷ t = 0,693 ÷ 5700</span> y = 0,000121 1/y = 1,21·10⁻⁴ y.
c = c₀·e∧-λ·t.
c = 2000 · e∧-(0,000121 1/y · 17100 y).
c = 252 g.
ω = 252 g ÷ 2000 g = 0,126 = 12,6%.
<span>
</span>
Answer:
The correct answers are "They travel slower than P waves", "They result in much ground motion" and " They are produced by P and S waves".
Explanation:
A surface wave can travel through the existing interface between Earth and air, as well as between Earth and water. A clear example of this type of waves are Love and Rayleigh waves. They travel slower than primary waves, produce large movements in the ground and are produced by primary and secondary waves.
Have a nice day!
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:
