Answer:
CH₃CH₂CH₂COOH > CH₃CH₂COOH > ClCH₂CH₂COOH > ClCH₂COOH
Explanation:
Electron-withdrawing groups (EWGs) increase acidity by inductive removal of electrons from the carboxyl group.
Electron-donating groups (EDGs) decrease acidity by inductive donation of electrons to the carboxyl group.
- The closer the substituent is to the carboxyl group, the greater is its effect.
- The more substituents, the greater the effect.
- The effect tails off rapidly and is almost zero after about three C-C bonds.
CH₃CH₂-CH₂COOH — EDG — weakest — pKₐ = 4.82
CH₃-CH₂COOH — reference — pKₐ = 4.75
ClCH₂-CH₂COOH — EWG on β-carbon— stronger — pKₐ = 4.00
ClCH₂COOH — EWG on α-carbon — strongest — pKₐ = 2.87
Answer:
31.9 °C
Explanation:
The formula for the heat q absorbed by an object is
q = mCΔT where ΔT = (T₂ - T₁)
Data:
q = 12.35 cal
m = 19.75 g
C = 0.125 cal°C⁻¹g⁻¹
T₂ = 37.0 °C
Calculations
(a) Calculate ΔT
q = mCΔT
12.35 cal = 19.25 g × 0.125 cal°C⁻¹g⁻¹ × ΔT
12.35 = 2.406ΔT °C⁻¹
ΔT = 12.35/(2.406 °C⁻¹) = 5.13 °C
(b) Calculate T₂
ΔT = T₂ - T₁
T₁ = T₂ - ΔT = 37.0 °C - 5.13 °C = 31.9 °C
The original temperature was 31.9 °C.
Answer:
The volume of HCl to be added to completely react with the ammonia is 0.032 L or 32mL
Explanation:
Using the formula
Ca Va = Cb Vb
Cb = 0.32 M
Vb = 50 mL = 50/1000 = 0.050L
Ca = 0.5 M
Va =?
Substituting for Va in the equation, we obtain:
Va = Cb Vb / Ca
Va = 0.32 * 0.05 / 0.5
Va = 0.016 / 0.5
Va = 0.032 L
The volume of HCl to be added to completely react with the ammonia is 0.032 L or 32mL
