Respuesta :
Answer:
1) 99.7 g is the total mass of the solution.
2)The amount of heat absorbed by solution is 2.00 kJ.
3) Moles of lithium nitrate are used is 0.036 mol.
4) The enthalpy for the dissolution reaction of one mole of lithium nitrate is -55.6 kJ/mol.
5) The value of q would be low.
Explanation:
1) mass of the water = M
Volume of the water,V = 97.2 mL
Density of the water ,d= 1.0 g/mL
[tex]Mass=Volume \times Density[/tex]
[tex]M=d\times V=1.0 g/ml\times 97.2 ml=97.2 g[/tex]
Mass of the lithium nitrate = M' = 2.5 g
Mass of the solution ,m = M + M' = 97.2 g + 2.5 g = 99.7 g
2) Mass of solution = m = 99.7 g
Specific heat of the solution = c = 4.184 J/g°C
Change in temperature = ΔT = 28.3°C - 23.5°C = 4.8°C
Temperature of the solution is increased which means that heat absorbed by the solution.
Heat energy absorbed by the solution = Q
[tex]Q=mc\Delta T[/tex]
[tex]=99.7 g\times 4.184 J/g^oC\times 4.8^oC=2,002.30 J=2.002 kJ\approx 2.00 kJ[/tex]
The amount of heat absorbed by solution is 2.00 kJ.
3) Mass of lithium nitrate = 2.5 grams
Molar mass of lithium nitrate = 69 g/mol
Moles of lithium nitrate , n= [tex]\frac{2.5 g}{69 g/mol}=0.036 mol[/tex]
Moles of lithium nitrate are used is 0.036 mol.
4) Moles of lithium nitrate = n = 0.036 mol
Heat absorbed by solution = heat released on dissolution = 2.00 kJ.
heat released on dissolution = -2.00 kJ
(negative sign indicates that heat is released)
Enthalpy of the dissolution reaction of one mole of lithum nitrate:
[tex]\Delta H_{dissolution}=\frac{-Q}{n}=\frac{2.00 kJ}{0.036 mol}=-55.6 kJ/mol[/tex]
The enthalpy for the dissolution reaction of one mole of lithium nitrate is -55.6 kJ/mol.
5) If the lid was left off during the temperature measurement the the heat absorbed by the solution will lost some amount of heat to the atmosphere which will increase the temperature the solution and will also effect the other observations.
