Answer:
9.4 liter
Explanation:
1) Data:
V₁ = 10.0 L
T₁ = 25°C = 25 + 273.15 K = 298.15 K
P₁ = 98.7 Kpa
T₂ = 20°C = 20 + 273.15 K = 293.15 K
P₂ = 102.7 KPa
V₂ = ?
2) Formula:
Used combined law of gases:
PV / T = constant
P₁V₁ / T₁ = P₂V₂ / T₂
3) Solution:
Solve the equation for V₂:
V₂ = P₁V₁ T₂ / (P₂ T₁)
Substitute and compuite:
V₂ = P₁V₁ T₂ / (P₂ T₁)
V₂ = 98.7 KPa × 10.0 L × 293.15 K / (102.7 KPa × 298.15 K)
V₂ = 9.4 liter ← answer
You can learn more about gas law problems reading this other answer on
Explanation:
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:
rate= k[A]²[B]²[C]
Explanation:
When concentration of A is increased two times ,keeping other's concentration constant , rate of reaction becomes 4 times .
So rate is proportional to [A]²
When concentration of B is increased two times , keeping other's concentration constant,rate of reaction becomes 4 times.
So rate is proportional to [B]²
When concentration of C is increased two times , keeping other's concentration constant, rate of reaction becomes 2 times.
So rate is proportional to [C]
So rate= k[A]²[B]²[C]
