Answer:
Of the following equilibria, only one will shift to the right in response to a decrease in volume.
On decreasing the volume the equilibrium will shift in right direction due to less number of gaseous moles on product side.
Explanation:
Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.
This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.
Decrease the volume
If the volume of the container is decreased , the pressure will increase according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease in pressure is taking place. So, the equilibrium will shift in the direction number of gaseous moles are less.
On decreasing the volume the equilibrium will shift in right direction due to less number of gaseous moles on product side.
On decreasing the volume the equilibrium will shift in left direction due to less number of gaseous moles on reactant side.
On decreasing the volume the equilibrium will shift in left direction due to less number of gaseous moles on reactant side.
On decreasing the volume the equilibrium will shift in no direction due to same number of gaseous moles on both sides.
On decreasing the volume the equilibrium will shift in no direction due to same number of gaseous moles on both sides.
Answer:
<h2>117.94 moles</h2>
Explanation:
To find the number of moles in a substance given it's number of entities we use the formula
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question we have
We have the final answer as
<h3>117.94 moles</h3>
Hope this helps you
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
This is the answer to question 19. If it’s not clear, send me a message.
Answer:
1.55 mol
Step-by-step explanation:
We can use the <em>Ideal Gas Law</em> to solve this problem.
pV = nRT Divide both sides by RT
n = (pV)/(RT)
Data:
p = 1.40 atm
V = 27.5 L
R = 0.082 06 L·atm·K⁻¹mol⁻¹
T = 302 K
Calculations:
n = (1.40 × 27.5)/(0.082 06 × 302)
= 1.55 mol
