Answer:
the Molar heat of Combustion of diphenylacetylene =
Explanation:
Given that:
mass of diphenylacetylene = 0.5297 g
Molar Mass of diphenylacetylene = 178.21 g/mol
Then number of moles of diphenylacetylene =
=
= 0.002972 mol
By applying the law of calorimeter;
Heat liberated by 0.002972 mole of diphenylacetylene = Heat absorbed by + Heat absorbed by the calorimeter
Heat liberated by 0.002972 mole of diphenylacetylene = msΔT + cΔT
= 1369 g × 4.184 J g⁻¹°C⁻¹ × (26.05 - 22.95)°C + 916.9 J/°C (26.05 - 22.95)°C
= 17756.48 J + 2842.39 J
= 20598.87 J
Heat liberated by 0.002972 mole of diphenylacetylene = 20598.87 J
Heat liberated by 1 mole of diphenylacetylene will be =
= 6930979.139 J/mol
= 6930.98 kJ/mol
Since heat is liberated ; Then, the Molar heat of Combustion of diphenylacetylene =
Answer:
A. have subjects wear the shoes in an experimental trial to see if symptoms improve.
The mass of nitrogen collected is mathematically given as
M-N2=0.025gram
<h3>What is the mass of nitrogen collected?</h3>
Question Parameters:
A sample weighing 2.000g
the liberated NH3 is caught in 50ml pipeful of H2SO4 (1.000ml = 0.01860g Na2O).
T=26.3c=299.3K
Pressure=745mmHg=745torr
Pressure of N2=745-25.2=719.8torr
Generally, the equation for the ideal gas is mathematically given as
PV=nRT
Therefore
719.8/760=45.6/1000=n*0.0821*299.3
n=0.00176*14
In conclusion, the Mass of N2
M-N2=0.00176*14
M-N2=0.025gram
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Answer:
0.200 m K3PO3
Explanation:
Let us remember that the freezing point depression is obtained from the formula;
ΔTf = Kf m i
Where;
Kf = freezing point constant
m = molality
i = Van't Hoff factor
The Van't Hoff factor has to do with the number of particles in solution. Let us consider the Van't Hoff factor for each specie.
0.200 m HOCH2CH2OH - 1
0.200 m Ba(NO3)2 - 3
0.200 m K3PO3 - 4
0.200 m Ca(CIO4)2 - 3
Hence, 0.200 m K3PO3 has the greatest van't Hoff factor and consequently the greatest freezing point depression.