When waters salinity increases it’s freezing point will decrease. This is one of the reasons why people salt the roads after a snow storm.
Answer:
C. CH3COOH, Ka = 1.8 E-5
Explanation:
analyzing the pKa of the given acids:
∴ pKa = - Log Ka
A. pKa = - Log (1.0 E-3 ) = 3
B. pKa = - Log (2.9 E-4) = 3.54
C. pKa = - Log (1.8 E-5) = 4.745
D. pKa = - Log (4.0 E-6) = 5.397
E. pKa = - Log (2.3 E-9) = 8.638
We choose the (C) acid since its pKa close to the expected pH.
⇒ For a buffer solution formed from an acid and its respective salt, we have the equation Henderson-Hausselbach (H-H):
- pH = pKa + Log ([CH3COO-]/[CH3COOH])
∴ pH = 4.5
∴ pKa = 4.745
⇒ 4.5 = 4.745 + Log ([CH3COO-]/[CH3COOH])
⇒ - 0.245 = Log ([CH3COO-]/[CH3COOH])
⇒ 0.5692 = [CH3COO-]/[CH3COOH]
∴ Ka = 1.8 E-5 = ([H3O+].[CH3COO-])/[CH3COOH]
⇒ 1.8 E-5 = [H3O+](0.5692)
⇒ [H3O+] = 3.1623 E-5 M
⇒ pH = - Log ( 3.1623 E-5 ) = 4.5
Explanation:
Zn=65
Cl2= 35+35=70
65+70=135g
1 mole ZnCl2 = 135g
x mole = 17.5g
17.5g × 1 mole/ 135g= 0.129 moles en 17.5g de ZnCl2
1. 2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂
2. CH₄ + 2O₂ → CO₂ + 2H₂O
3. Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag
4. MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂
5. Pb(NO₃)₂ + 2NaCl → PbCl₂ + 2NaNO₃
1)

2)
CuSO_4+Cu_2Cl_2\neq>

They both break down and and erode