Chemistry chapter 8 Entry Test MCQs

a. Na+(g) +I-(g)———>NaI(s)
b. K+(g) + Cl(g)——->KCl(s)
c. Na+(g) +Cl(g)———>NaCl(s)
d. K+(g) + Br(g)———>KBr(aq)

d. K+(g) + Br(g)———>KBr(aq)

a. Enthalpy
b. Internal energy
c. Temperature
d. Volume

a. Enthalpy

a. Order of the reaction
b. Mechanism of the reaction
c. Stoichiometry of the reaction
d. Heat of the reaction

d. Heat of the reaction

a. bomb calorimeter
b. glass calorimeter
c. manometer
d. thermometer

a. bomb calorimeter

a. heat of combustion
b. heat of solution
c. heat of formation of H2O
d. heat of atomization

c. heat of formation of H2O

a. -273 0C and 1 atm
b. 273 K and 1 atm
c. 25 0C and 1 atm
d. 0 0C and 1 atm

c. 25 0C and 1 atm

a. 2.79 J
b. 627 kJ
c. 627 J
d. 6.27 KJ

c. 627 J

a. Twice of x
b. More than x
c. Less than x
d. Equal to x

c. Less than x

a. 6.4 kg
b. 9.6 kg
c. 2.8 kg
d. 96.0 kg

b. 9.6 kg

a. When 1 mole of Reactants are converted into products
b. When 1 mol of compound formed from its elements
c. When 1 mole of salt is formed
d. When 1 mole of substance is dissolved to make a solution

a. When 1 mole of Reactants are converted into products

a. 16.2 Kj/mol
b. Kj/mol
c. -16.2 Kj/mol
d. -17 Kj/mol

a. 16.2 Kj/mol

a. q=m x v x ΔT
b. q=m x s x ΔT
c. q = pv
d. q=ΔE

b. q=m x s x ΔT

Hess`s Law

a. Law of conservation of energy
b. Law of Mass Action
c. Dalton’s Law of partial pressure
d. Law of conservation of mass

a. Law of conservation of energy

a. it reacts with CO2
b. it does not catch fire
c. It is not prepared directly
d. protective layer of oxide cover the surface

d. protective layer of oxide cover the surface

a. > 40 KJ
b. 40 kJ
c. Zero
d. < 40 Kj

c. Zero

a. Law of mass action
b. First law of thermodynamics
c. Raoult’s law
d. Hess’s law

d. Hess’s law

a. Formation
b. Reaction
c. Transition
d. All of these

a. Formation

a. Initial reactants and final products
b. Initial reactants only
c. Intermediates only
d. Final products only

a. Initial reactants and final products

a. CO
b. CCl4
c. C
d. O3

a. CO

a. The lattice energy of crystalline substances can be calculated easily
b. Born Harbor cycle is different from Hess’s law
c. It is applied for M2X compounds
d. The energy changes in a cyclic process is not zero

a. The lattice energy of crystalline substances can be calculated easily

a. completely in oxygen
b. completely in air
c. with nitrogen
d. with carbon

a. completely in oxygen

a. MgF2
b. KF
c. CaF2
d. NaF

a. MgF2

a. Heat of hydration
b. Heat energy
c. Heat of solution
d. Lattice energy

d. Lattice energy

a. Properties
b. Structure
c. Dipole Moment
d. Bonding

c. Dipole Moment

a. second law of thermodynamics
b. first law of thermodynamics
c. Hess’s law
d. first law of thermochemistry

c. Hess’s law

Similar Posts

Leave a Reply

Your email address will not be published. Required fields are marked *