B. Low blood pressure and drooling. If the user is tranquilized, he would not be able to hinder the production of saliva, leading to drooling. And if the victim was tranquilized, his heart would slow down, not speed up, resulting in a lower blood pressure.
Answer:
v = 2,66x10⁻⁵ P[H₂C₂O₄]
Explanation:
For the reaction:
H₂C₂O₄(g) → CO₂(g) + HCOOH(g)
At t = 0, the initial pressure is just of H₂C₂O₄(g). At t= 20000 s, pressures will be:
H₂C₂O₄(g) = P₀ - x
CO₂(g) = x
HCOOH(g) = x
P at t=20000 is:
P₀ - x + x + x = P₀+x. That means P at t=20000s - P₀ = x
For 1st point:
x = 92,8-65,8 = 27
Pressure of H₂C₂O₄(g) at t=20000s: 65,8-27 = 38,8
2nd point:
x = 130-92,1 = 37,9
H₂C₂O₄(g): 92,1 - 37,9 = 54,2
3rd point:
x = 157-111 = 46
H₂C₂O₄(g): 111-46 = 65
Now, as the rate law is :
v = k P[H₂C₂O₄]
Based on integrated rate law, k is:
(- ln P[H₂C₂O₄] + ln P[H₂C₂O₄]₀) / t = k
1st point:
k = 2,64x10⁻⁵
2nd point:
k = 2,65x10⁻⁵
3rd point:
k = 2,68x10⁻⁵
The averrage of this values is:
k = 2,66x10⁻⁵
That means law is:
v = 2,66x10⁻⁵ P[H₂C₂O₄]
I hope it helps!
Explanation:
Formula to calculate work done by motor is as follows.
Work done by motor = 
where, g = gravitational constant = 10 
Therefore, work done by motor is as follows.
Work done by motor = 
= 100.0 J
Now, heat lost by water will be calculated as follows.
q = 
= 
= 10.0 J
Hence, heat gained by motor = heat lost by water
As, heat gained by motor = 10.0 J
So, change in energy = heat gained - work done
Therefore, change in energy will be calculated as follows.
Change in energy = heat gained - work done
= (10.0 J) - (100.0 J)
= -90.0 J
Thus, we can conclude that change in the energy of the battery contents is -90.0 J.
<span>Friction creates heat which in turn can lead to deviations from the original size and shape of a part.
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Answer:
25 possibly
Explanation:
I'm not too sure about this, but sodium oxide is Na2O, 2 sodium and 1 oxygen, so 12.5g * 2 is 25
If someone else comes up with a more convincing argument listen to them
