Answer:
a) pH = 4.213
b) % dis = 2 %
Explanation:
Ch3COONa → CH3COO- + Na+
CH3COOH ↔ CH3COO- + H3O+
∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]
mass balance:
⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [ CH3COOH ] + [ CH3COO- ]
<em>∴ C </em>CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)
∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL
⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]
charge balance:
⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water
⇒ [ CH3COO- ] = [ H3O+ ] + 1.00
⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5
⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]
⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0
⇒ [ H3O+ ] = 6.12 E-5 M
⇒ pH = - Log [ H3O+ ] = 4.213
b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4
∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol
⇒ % dis = 3.4 / 1.7 = 2 %
Answer:
The covalent bond in Cl₂ is break and combine with sodium to form NaCl through ionic bond.
Explanation:
Chemical equation:
Na + Cl₂ → NaCl
Balanced chemical equation:
2Na + Cl₂ → 2NaCl
The given reaction indicate the formation of sodium chloride.
Sodium chloride is an ionic compound. It is formed by the reaction of chlorine and sodium. The type of bond in Cl₂ is covalent. Both chlorine atoms are tightly held together through sharing of electrons. When sodium chloride is formed the covalent between the chlorine atoms are break and it react with sodium . The chlorine toms thus gain the one electron from the sodium atom and became negative ion while sodium by losing its one valance electrons became positive ions. The strong electrostatic forces are develop between them and ionic bond is formed.
I am sure, the answer is variant B.
Answer: We do not know. We have not been given the solubility of oxygen in water at a given temperature nor have we been given the Henry's laws constant. We also do not know whether you mean 1 atmosphere of air, or 1 atmosphere of oxygen.
Answer:
Na + CaSO4 = Na2SO4 + Ca
Explanation:
single displacement (substitution)
