Suppose that a TLC plate is spotted with samples 1.0 cm above the bottom of the plate. Select the consequence of inserting the p
1 answer:
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We need to use the following formula
Δ
n= 4 moles
F= constant= 96500C/mol
let's plug in the values.
ΔG= -(4)(96500)(0.24)= -92640 J or -92.6 kJ
Δ
n= 4 moles
F= constant= 96500C/mol
let's plug in the values.
ΔG= -(4)(96500)(0.24)= -92640 J or -92.6 kJ
Answer:
The total pressure of the mixture in the tank of volume 6.25 litres at 51°C is 1291.85 kPa.
Explanation:
For N2,
Pressure(P₁)=125 kPa
Volume(V₁)=15·1 L
Temperature (T₁)=25°C=25+273 K=298 K
Similarly, for Oxygen,
Pressure(P₂)= 125 kPa
Volume(V₂)= 44.3 L
Temperature(T₂)=25°C= 298 K
Then, for the mixture,
Volumeof the mixture( V)= 6.25 L
Pressure(P)=?
Temperature (T)= 51°C = 51+273 K=324 K
Then, By Combined gas laws,
or,
or,
or,
∴P=1291.85 kPa
So the total pressure of the mixture in the tank of volume 6.25 litres at 51°C is 1291.85 kPa.
25.9 kJ/mol. (3 sig. fig. as in the heat capacity.)
<h3>Explanation</h3>The process:
.
How many moles of this process?
Relative atomic mass from a modern periodic table:
- K: 39.098;
- N: 14.007;
- O: 15.999.
Molar mass of :
.
Number of moles of the process = Number of moles of dissolved:
.
What's the enthalpy change of this process?
for
. By convention, the enthalpy change
measures the energy change for each mole of a process.
.
The heat capacity is the least accurate number in these calculation. It comes with three significant figures. As a result, round the final result to three significant figures. However, make sure you keep at least one additional figure to minimize the risk of rounding errors during the calculation.