HC₂H₃O₂ , equilibrium can be represented as:
HC₂H₃O₂ + H₂O--------->CH₃COO⁻ + H₃O⁺
HC₂H₃O₂, that is acetic acid dissociate in water that is H₂O and formed acetate anion that is CH₃COO⁻ and hydronium ions that is H₃O⁺.
Its equiibrium constant can be represented as:
Ka=\frac{[CH_{3}COO^{-1}]\times [H_{3}O^{+1}]}{CH_{3}COOH}
While NaC₂H₃O₂ can be dissociate as:
NaC₂H₃O₂------>CH₃COO⁻ + Na⁺
Sodium acetate that is NaC₂H₃O₂, dissociates as forming acetate anion and sodium cation.
CH4 (g) + 2Cl2 (g) --> CH2Cl2 (g) + 2HCl (g) <span>ΔH = -205.4</span>
Answer:
Option D. 30 g
Explanation:
The balanced equation for the reaction is given below:
2Na + S —> Na₂S
Next, we shall determine the masses of Na and S that reacted from the balanced equation. This is can be obtained as:
Molar mass of Na = 23 g/mol
Mass of Na from the balanced equation = 2 × 23 = 46 g
Molar mass of S = 32 g/mol
Mass of S from the balanced equation = 1 × 32 = 32 g
SUMMARY:
From the balanced equation above,
46 g of Na reacted with 32 g of S.
Finally, we shall determine the mass sulphur, S needed to react with 43 g of sodium, Na. This can be obtained as follow:
From the balanced equation above,
46 g of Na reacted with 32 g of S.
Therefore, 43 g of Na will react with = (43 × 32)/46 = 30 g of S.
Thus, 30 g of S is needed for the reaction.
Answer:el número de masa es la suma de los protones y neutrones en un átomo y el número atómico es sólo el número de protones.
Explanation: