The two substances have different densities. Density can be affected by the temperature of a substance. Since they have to same volume but weigh differently, they have different densities. Remember, density = mass/volume
Answer:
-3.7771 × 10² kJ/mol
Explanation:
Let's consider the following equation.
3 Mg(s) + 2 Al³⁺(aq) ⇌ 3 Mg²⁺(aq) + 2 Al(s)
We can calculate the standard Gibbs free energy (ΔG°) using the following expression.
ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)
where,
n: moles
ΔG°f(): standard Gibbs free energy of formation
p: products
r: reactants
ΔG° = 3 mol × ΔG°f(Mg²⁺(aq)) + 2 mol × ΔG°f(Al(s)) - 3 mol × ΔG°f(Mg(s)) - 2 mol × ΔG°f(Al³⁺(aq))
ΔG° = 3 mol × (-456.35 kJ/mol) + 2 mol × 0 kJ/mol - 3 mol × 0 kJ/mol - 2 mol × (-495.67 kJ/mol)
ΔG° = -377.71 kJ = -3.7771 × 10² kJ
This is the standard Gibbs free energy per mole of reaction.
Answer: The answer is 68142.4 Pa
Explanation:
Given that the initial properties of the cylindrical tank are :
Volume V1= 0.750m3
Temperature T1= 27C
Pressure P1 =7.5*10^3 Pa= 7500Pa
Final properties of the tank after decrease in volume and increase in temperature :
Volume V2 =0.480m3
Temperature T2 = 157C
Pressure P2 =?
Applying the gas law equation (Charles and Boyle's laws combined)
P1V1/T1 = P2V2/T2
(7500 * 0.750)/27 =( P2 * 0.480)/157
P2 =(7500 * 0.750* 157) / (0.480 *27)
P2 = 883125/12.96
P2 = 68142.4Pa
Therefore the pressure of the cylindrical tank after decrease in volume and increase in temperature is 68142.4Pa
Q1)
the number of moles can be calculated as follows
number of moles = mass present / molar mass
number of moles is the amount of substance.
4.8 g of Ca was added therefore mass present of Ca is 4.8 g
molar mass of Ca is 40 g/mol
molar mass is the mass of 1 mol of Ca
therefore if we substitute these values in the equation
number of moles of Ca = 4.8 g / 40 g/mol = 0.12 mol
0.12 mol of Ca is present
q2)
next we are asked to calculate the number of moles of water present
again we can use the same equation to find the number of moles of water
number of moles = mass present / molar mass
3.6 g of water is present
sum of the products of the molar masses of the individual elements by the number of atoms
H - 1 g/mol and O - 16 g/mol
molar mass of water = (1 g/mol x 2 ) + 16 g/mol = 18 g/mol
molar mass of H₂O is 18 g/mol
therefore number of moles of water = 3.6 g / 18 g/mol = 0.2 mol
0.2 mol of water is present
