A lead–tin alloy of composition 30 wt% Sn–70 wt% Pb is slowly heated from a temperature of 150°C (300°F).(a) At what temperature
1 answer:
Answer:
a) 231.9 °C
b) 100% Sn
c) 327.5 °C
d) 100% Pb
Explanation:
This is a mixture of two solids with different fusion point:
<u>Given that Sn has a lower fusion temperature it will start to melt first at that temperature. </u>
So the first liquid phase forms at 231.9 °C and because Pb starts melting at a higher temperature, that phase's composition will be 100% Sn.
The mixture will be completely melted when you are a the higher melting temperature of all components (in this case Pb), so it will all in liquid phase at 327.5 °C.
At that temperature all Sn was already in liquid state and, therefore, the last solid's composition will be 100% Pb.
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Answer:
The molarity of this solution is 0,09254M
Explanation:
The concentration of the Ni²⁺ solution is:
Ni²⁺ + EDTA⁴⁻ → Ni(EDTA)²⁻
0,03935L × 0,01307M = 5,143x10⁻⁴ moles Ni²⁺ ÷ 0,03010L =<em>0,01709M Ni²⁺</em>
25,00 mL of this solution contain:
0,01709M × 0,02500L = 4,2716x10⁻⁴ moles of Ni²⁺
The moles of Ni²⁺ that are in excess and react with EDTA⁴⁻ are:
0,01015L × 0,01307M = 1,3266x10⁻⁴ moles of Ni²⁺
Thus, moles of Ni²⁺ that react with CN⁻ are:
4,2716x10⁻⁴ - 1,3266x10⁻⁴ = 2,9450x10⁻⁴ moles of Ni²⁺
For the reaction:
4CN⁻ + Ni²⁺ → Ni(CN)₄²⁻
Four moles of CN⁻ react with 1 mole of Ni²⁺:
2,9450x10⁻⁴ moles of Ni²⁺ × = <em>1,178x10⁻³ moles of CN⁻</em>
As the volume of cyanide solution is 12,73mL. The molarity of this solution is:
<em>1,178x10⁻³ moles of CN⁻ ÷ 0,01273L = </em><em>0,09254M</em>
I hope it helps!