The composition, in atom percent, of an alloy that consists of 30 wt% Zn and 70 wt% Cu is 29.4 at % and 70.6 at % for Zn and Cu respectively.
<h3>How to convert weight % to atom %?</h3>
The weight percentage of an element can be converted to atom percentage as follows:
- C(Zn) = C1A2 / (C2A1) + (C1A2) × 100
- C(Cu) = C2A1 / (C1A2) + (C2A1) × 100
Where;
- C1 = 30
- C2 = 70
- A1 = 65.4g/mol
- A2 = 63.55g/mol
- C(Zn) = {30*63.55} / (70*65.4) + (30*63.55) × 100
- C(Cu) = {70*65.4} / (30*63.55) + (70*65.4) × 100
- C(Zn) = 29.4 at %
- C(Cu) = 70.6 at %
Therefore, the composition, in atom percent, of an alloy that consists of 30 wt% Zn and 70 wt% Cu is 29.4 at % and 70.6 at % for Zn and Cu respectively.
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When answering this question, you must consider the reactivity series. Mg is much higher than Zn in the reactivity series, therefore it can displace the Zn from the solution - displacement reaction.
It seems odd that you'd react Magnesium nitrate with zinc nitrate. Maybe you meant magnesium with zinc nitrate or vice Versa. Because in your case each compound would remain as it is as both Mg and Zn are happy in a relatively stable compound.
As for magnesium nitrate and magnesium nitrate, you cannot achieve a reaction between two substances that are the same. The same applies for Mg and magnesium nitrate. No displacement happens.
Hope it helps!
1)The third equation must be halved.
H20–>H2+1/2O2
So ΔΗ3=483.6/2
2)The second equation must be halved and reversed.
CO2–>CO+1/2O2
So ΔΗ2=-(-566)/2=566/2
And now :
C+O2–>CO2
CO2–>CO+1/2O2
H2O—>H2+1/2O2
(You add them)
C+H2O—>CO+H2
ΔΗrxn=ΔΗ1+ΔΗ2’+ΔΗ3’=-393.5+283+241,8=131,3
