Li2S + 2 HNO3 --> 2 LiNO3 + H2S
Li2 S + H2 N2 O2 --> Li2 N2 O5 + H2 S
Li S + H2 N2 O5 -> Li N2 O5 + H2 S
Li2 S2 + H4 N4 O10 --> Li2 N4 O10 + H4 S2
Li^2 S^2 + H^4 N^4 O^10 --> Li^2 N^4 O^10 + H^4 S^2
The value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).
Aa we know that, 125mL of 0.06M Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl.
Given, T = 25°C.
<h3>Chemical equation:</h3>
Pb(NO3)2 + NaCl ---- NaNO3 + PbCl2
PbCl2 in aqueous solution split into following ions
PbCl2 ------ Pb(+2) + 2Cl-
Q = [Pb(+2)] [Cl-]^2
The Concentration of Pb(+2) ions and Cl- ions can be calculated as
[Pb(+2)] = 0.06 × 125/200
= 0.0375
[Cl-] = 0.02 × 75/200
= 0.0075
By substituting all the values, we get
[0.0375] [0.0075]^2
= 2.11 × 10^(-6).
Thus, we calculated that the value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).
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Answer is: the approximate freezing point of a 0.10 m NaCl solution is -2x°C.
V<span>an't
Hoff factor (i) for NaCl solution is approximately 2.
</span>Van't Hoff factor (i) for glucose solution is 1.<span>
Change in freezing point from pure solvent to
solution: ΔT = i · Kf · m.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
m - molality, moles of solute per
kilogram of solvent.
</span>Kf and molality for this two solutions are the same, but Van't Hoff factor for sodium chloride is twice bigger, so freezing point is twice bigger.
1 kPa = 7.5 mmHg so 7.0 mmHg / 7.5 mmHg x 1 kPa = .93 kPa
101.3 kPa = 1 atm so 10 kPa / 101.3 kPa x 1 atm = .0987 atm
1 kPa = 7.5 mmHg so 15 kPa x 7.5 mmHg / 1 kPa = 112.5 mmHg
