The mathematical expression for heat capacity at constant pressure is given as:
(1)
where, Q = heat capacity
= molar heat capacity at constant pressure
= change in temperature
n = number of moles
Therefore, = 
= 
Number of moles =
= 
= 0.186 mole
Put the values in formula (1)
(conversion of degree Celsius into kelvin)
= 
= 6.32 J /mol K
Hence, molar heat capacity of benzene at constant pressure = 
Answer: Mass of zinc that reacts with 4.11 g of hydrochloric acid to form 9.1 g of zinc chloride and 3.97 g of hydrogen gas is 8.96 g
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
Given: mass of hydrochloric acid = 4.11 g
Mass of products = Mass of zinc chloride + mass of hydrogen = 9.1 g + 3.97 g = 13.07 g
As mass of reactant = mass of products
mass of hydrochloric acid + mass of zinc = Mass of zinc chloride + mass of hydrogen
4.11 g + mass of zinc = 13.07 g
mass of zinc = 8.96 g
Answer:
H2 P4 O1. Explanation: In order to calculate the Empirical formula , we will assume that we have started with 10 g of the compound.
Explanation:
Answer:option d==> Si.
Explanation:
The energy required to remove electron from a gaseous atom or ion is what is called an ionization energy. As we remove electrons continually in a gaseous atom or ion, the ionization energy increases which are know as the first ionization energy, the second ionization energy, third ionization energy and so on.
Looking at the electronic configuration of Silicon, Si; Ne 3s2 3p2. We can can see that the first four ionization energies are from the removal of the 3p2 and 3s2 electrons and the fifth ionization energy, which is the highest ionization energy of 14800 kJ/mol is the the electron removed from the core shell.
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.
