Answer:
a) The mass of lead in the alloy = 0.28 kg
b) There can be dissolved 1.46 kg more of lead.
Explanation:
Step 1: Data given
mass of the magnesium-lead alloy = 5.5 kg
200 °C → 5 wt% Pb
350 °C → 25 wt% Pb
a) What mass of lead is in the alloy?
We have to calculate (for the magnesium-lead alloy) the mass of lead in 5.5 kg of the solid α phase at 200°C just below the solubility limit. The solubility limit for the α phase at 200°C is about 5 wt% Pb.
The mass of lead in the alloy = (0.05)*(5.5 kg) = 0.28 kg
b) If the alloy is heated to 350°C, how much more lead may be dissolved in the α phase without exceeding the solubility limit of this phase?
At 350°C, the solubility limit of the a phase increases to approximately 25 wt% Pb.
C(lead) = ((mass fo lead in alloy) + (mass lead)) / ((magnesium-lead alloy mass) + (mass lead))
0.25 = (0.28 + m(Pb)) / ( 5.5 + m(Pb))
0.25 * ( 5.5 + m(Pb)) = (0.28 + m(Pb))
1.375 + 0.25 m(Pb) = 0.28 + m(Pb)
1.095 = 0.75 m(Pb)
m(Pb) = 1.095 / 0.75
m(Pb) = 1.46
There can be dissolved 1.46 kg more of lead.
Fluid illustrations incorporate unadulterated water, white vinegar, sugar water, corn oil, and blood plasma. Shockingly, homogeneous blends are not restricted to fluids, they can likewise be gases and solids. Another kind of homogeneous blend is known as a colloid.
When a solution of Pb(NO3)2(aq) is mixed with a solution of KI(aq), a precipitate of PbI₂ will form; K⁺ and NO₃⁻ are spectator ions.
<u>Explanation:</u>
When an aqueous solution of lead nitrate (Pb(NO₃)₂ is mixed with aqueous solution of potassium iodide (KI), then there is a precipitate formation of lead iodide (PbI₂), and the potassium (K⁺) ion and nitrate (NO₃⁻) ion acts as spectator ions that is ions do not involved in the reaction.
The reaction can be represented as,
Pb(NO₃)₂(aq) + 2 KI (aq) → PbI₂(s) + 2KNO₃(aq)
The ionic equation can be written as,
Pb²⁺(aq) + 2 NO₃⁻(aq) + 2 K⁺(aq) + 2 I⁻(aq) → PbI₂(s) + 2K⁺(aq) + 2NO₃⁻(aq)
On both sides of the equation, we have K⁺ and NO₃⁻ ions, which gets cancelled, and these 2 ions are called as spectator ions.
All the answers in the equation above will be Zero (0)
We let x be the pressure of each product at equilibrium, giving this ICE table:
2NO2 (g) ↔ 2NO (g) + O2 (g)
Initial pressure (atm): 0.500 0 0
Change (atm): -2x +2x +x
Equilibrium (atm): 0.500-2x 2x x
We can calculate x from Kp:
Kp = [NO]^2 [O2] / [NO2]^26.5x10^-6 = (2x)^2 (x) / (0.500-2x)^2
Approximating that 2x is negligible compared to 0.500 simplifies the equation to
6.5x10^-6 = (2x)^2 (x)/(0.500)^2 = 4x^3 / (0.500)^2
Then we solve for x:
x3 = (6.5x10^-6)(0.500)^2 / 4
x = 0.00741
The pressure of NO2 at equilibrium is therefore
Pressure of NO2 = 0.500-2x = 0.500 - 2(0.00741) = 0.4852 atm