If molecule a has a bond enthalpy of 418 kj/mol and molecule b has a bond enthalpy of 941 kj/mol, what can be said of molecule a compared to molecule b

Options for the question.

A. It requires less energy to break the bonds in molecule A than it does in molecule B.

B.It requires more energy to break the bonds in molecule A than it does in molecule B.

C. Molecule A is more stable than molecule B.

D. Molecule A has stronger bonds than molecule B.

A. It requires less energy to break the bonds in molecule A than it does in molecule B.

Explanation:

In chemistry, bond energy (E) or bond enthalpy (H) is the measure of bond strength in a chemical bond.

The higher the bond enthalpy, the more energy is needed to break the bond and the stronger the bond. The lower the bond enthalpy, the lesser energy is needed to break the bond and the weaker the bond.

The correct option is A. Since A has a lower enery value compared to B, it would take a lesser amount of energy to break the bonds in A.

Elimination of other options;

B) B has more bond enthalpy so more energy is needed hence this option is wrong.

C) Bond energies (enthalpies) can be used to indicate how stable a compound is or how easy it is to break a particular bond. The more energy that is required to break a bond, the more stable the compound will be. A larger bond energy implies the bond is harder to break, so the compound will be more stable. B would be more stable hence this option is wrong.

D) Bond energy is proportional to strength of bonds. Hence this option is also wrong.

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.