First, we have to get moles of CH3COONa = mass/molar mass
= 20 g / 82.03 g/mol = 0.244 moles
when we have [CH3COOH] = 0.15 M
∴ [CH3COONa] = moles of CH3COONa / Volume of solution
= 0.244 moles / 0.5L = 0.488 M
when we look up for Ka of acetic acid value it is equal 1.8 x 10^-5
So we can get Pka = -㏒Ka
= -㏒(1.8 x10^-5)
= 4.7
now we will use Henderson - Hasselbalchn equation to get the PH:
PH = Pka + ㏒[conjugate basic/weak acid]
when CH3COOH is the weak acid & CH3COO- is the conjugate base so by substitution:
PH = 4.7 + ㏒ (0.488/0.15)
= 5.2
b) when we have this equation for the reaction:
HCl + CH3COONa → CH3COOH + NaCl
ionic equation : H+ + Cl- + CH3COO- + Na+ → CH3COOH + Na+ + Cl-
when HCl + H2O → H3O+ + Cl-
∴ the reaction will be:
CH3COO- (aq) + H3O+(aq) → CH3COOH(aq) + H2O(l)
Answer:
4.62
so 5
the ratio is 2 na chlorates for 3 O2 so multiply 7 by 2/3
Explanation:
Answer:
Reactions, 2, 3 and 5 make precipitates
Explanation:
1. The halogens always make aqueous salts with elements from group 1
2. Phosphate anion can make insoluble salts, the same as carbonate.
3. Nitrate anion always make aqueous salts
4. Sulfate anion makes aqueous salts except with Ag⁺, Pb⁺² and group 2
1. KI(aq) + NaCl(aq) → KCl(aq) + NaI(aq)
2. 2Na₃PO₄ (aq) + 3CoCl₂(aq) → 6NaCl(aq) + Co₃(PO₄)₂(s) ↓
3. Na₂CO₃ (aq) + CuCl₂ (aq) → 2NaCl (aq) + CuCO₃ (s) ↓
4. 2LiNO₃ (aq) + Na₂SO₄ (aq) → Li₂SO₄ (aq) + 2NaNO₃(aq)
5. CrCl₂ (aq) + Li₂CO₃ (aq) → 2LiCl (aq) + Cr₂(CO₃)₂ (s) ↓
C
I have had this question on a test before!! Hope this helps
Answer is: solution of electrolyte will have lower freezing point than solution of nonelectrolyte.
This is because salt solution has more particles in of sodium chloride (sodium and chlorine ions) than in same concentration of glucose. Electrolytes better separates into particles in water because of their ionic bond.<span>
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