Answer:
δ N2(g) = 1.1825 g/L
Explanation:
- δ ≡ m/v
- Mw N2(g) = 28.0134 g/mol
ideal gas:
∴ P = (837 torr)×( atm/760 torr) = 1.1013 atm
∴ T = 45.0 °C + 273.15 = 318.15 K
∴ R = 0.082 atm.L/K.mol
⇒ n/V = P/R.T
⇒ n/V = (1.1013 atm) / ((0.082 atm.L/K.mol)(318.15 k))
⇒ n/V = 0.0422 mol/L
⇒ δ N2(g) = (0.042 mol/L)×(28.0134 g/mol) = 1.1825 g/L
<span> 52.0ml of 0.35M CH3COOH : 0.052 L(0.35M) = .0182 mol of CH3COOH.
</span>
<span>31.0ml of 0.40M NaOH : .031 L(0.40M) = .0124 mol of NaOH.
</span>
<span>After the reaction, .0124 Mol CH3COO- is generated and .058 mol CH3COOH is left un-reacted. The concentration would be 12.4/V and 5.8/V, respectively. Therefore:
</span>
<span>pH = -log([H+]) = -log(Ka*[CH3COOH]/[CH3COO-]) </span>
<span>= -log(1.8x10^-5*5.8/12.4) = 5.07</span>
Answer:
The conductance will increase as the concentration of the electrolyte is increased.
Explanation:
The ions are what carry the charges from one electrode to another. The more there are, the easier it is for electrons to get across the solution of electrolyte,
I’m soooo confusing with what your question is lol
Answer:
Silver
Explanation:
I remember this because Ag can mean "Ain't Gold"
