Answer:
<h2>The answer is 1.48 L</h2>
Explanation:
In order to find the original volume we use the same for Boyle's law which is
where
P1 is the initial pressure
P2 is the final pressure
V1 is the initial volume
V2 is the final volume
Since we are finding the original volume
From the question
P1 = 172 kPa = 172000 Pa
P2 = 85 kPa = 85000 Pa
V2 = 3 L
We have
We have the final answer as
<h3>1.48 L</h3>
Hope this helps you
First, we have to get the initial [C6H8O6] = mass/molar mass
when the molar mass of C6H8O6 = 176.12 g/mol
∴[C6H8O6] = 0.25 g / 176.12 g/mol
= 0.00142 M
when
C6H8O6 ⇄ H+ + C6H7O6-
intial 0.00142 M 0 0
change -X +X +X
Equ (0.00142-X) X X
so, Ka = [H+][C6H7O6-] / [C6H8O6]
by substitution:
8 x 10^-5 = X * X / (0.00142-X) by solving this equation for X
∴ X = 0.000299
∴[H+] = 0.000299
∴PH = -㏒[H+]
= -㏒ 0.000299
= 3.52
Answer:
1.73 atm
Explanation:
Given data:
Initial volume of helium = 5.00 L
Final volume of helium = 12.0 L
Final pressure = 0.720 atm
Initial pressure = ?
Solution:
"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"
Mathematical expression:
P₁V₁ = P₂V₂
P₁ = Initial pressure
V₁ = initial volume
P₂ = final pressure
V₂ = final volume
Now we will put the values in formula,
P₁V₁ = P₂V₂
P₁ × 5.00 L = 0.720 atm × 12.0 L
P₁ = 8.64 atm. L/5 L
P₁ = 1.73 atm
First, we write the half equations for the reduction of the chemical species present:
Cu⁺² + 2e → Cu; E° = 0.34 V
Ni⁺² + 2e → Ni; E° = - 0.23 V
In order to determine the potential of the cell, we find the difference between the two values. For this:
E(cell) = 0.34 - (-0.23)
E(cell) = 0.57 V
The second option is correct. (The difference in values is due to different values in literature, and it is negligible)
The more numbers after the decimal point there are, the more precise the instrument which recorded it is. For example, if one instrument during seismic activity records that the magnitude of the earthquake was 2.3, and another instrument recorded that it was 2.3645, the second instrument would have shown to be more precise.
