Answer:
pH = 12.15
Explanation:
To determine the pH of the HCl and KOH mixture, we need to know that the reaction is a neutralization type.
HCl + KOH → H₂O + KCl
We need to determine the moles of each compound
M = mmol / V (mL) → 30 mL . 0.10 M = 3 mmoles of HCl
M = mmol / V (mL) → 40 mL . 0.10 M = 4 mmoles of KOH
The base is in excess, so the HCl will completely react and we would produce the same mmoles of KCl
HCl + KOH → H₂O + KCl
3 m 4 m -
1 m 3 m
As the KCl is a neutral salt, it does not have any effect on the pH, so the pH will be affected, by the strong base.
1 mmol of KOH has 1 mmol of OH⁻, so the [OH⁻] will be 1 mmol / Tot volume
[OH⁻] 1 mmol / 70 mL = 0.014285 M
- log [OH⁻] = 1.85 → pH = 14 - pOH → 14 - 1.85 = 12.15
Answer: the basic difference is Exergonic reactions release energy and an endergonic reactions absorb energy .
HOPE THIS HELPS!!!
Answer:
Here's what I get.
Explanation:
- If your teachers don't ask for a specific type of formula, a condensed structural formula should be OK.
- If they ask specifically for a structural formula or a bond-line formula, that is what you must give.
Bottom line: ask your teachers in advance what they expect.
Answer:
∆H > 0
∆Srxn <0
∆G >0
∆Suniverse <0
Explanation:
We are informed that the reaction is endothermic. An endothermic reaction is one in which energy is absorbed hence ∆H is positive at all temperatures.
Similarly, absorption of energy leads to a decrease in entropy of the reaction system. Hence the change in entropy of the reaction ∆Sreaction is negative at all temperatures.
The change in free energy for the reaction is positive at all temperatures since ∆S reaction is negative then from ∆G= ∆H - T∆S, we see that given the positive value of ∆H, ∆G must always return a positive value at all temperatures.
Since entropy of the surrounding= - ∆H/T, given that ∆H is positive, ∆S surrounding will be negative at all temperatures. This is so because an endothermic reaction causes the surrounding to cool down.
The balanced equation
for the reaction is
CO(g) + 2H₂(g) ⇄ CH₃OH(g)
The given
concentrations are at equilibrium state. Hence we can use them directly in
calculation with the expression for the equilibrium constant, k.
expression for k can be written as
k = [CH₃OH(g)] / [CO(g)] [H₂<span>(g) ]²
</span>[H₂<span>]=0.072 M
[CO]= 0.020M
[CH</span>₃OH]= 0.030 M
From substitution,
k = 0.030
M / 0.020 M x (0.072 M)²
k =
289.35 M⁻²
<span>
Hence, equilibrium constant for the given reaction at 700 K is 289.35 M</span>⁻².
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