Consider the substances hydrogen (H2), fluorine (F2), and hydrogen fluoride (HF). Based on their molecular structures, how does
the boiling point of HF compare with the boiling points of H2 and F2? The boiling point of HF is (higher than, lower than, similar to) the boiling point of H2, and it is (higher than, lower than, similar to) the boiling point of F2.
The boiling point of HF is <u><em>higher than</em></u> the boiling point of H2, and it is <u><em>higher than</em></u> the boiling point of F2.
Explanation:
In HF, inter- molecule forces will be present between the hydrogen and fluorine atoms. There will be hydrogen bonding present among the hydrogen and fluorine atoms. Hydrogen bonds are strong bonds and hence the boiling point for HF would be high as much energy will be required to break these bonds.
H2 and F2 will only have intra-molecular attractions and there will be no hydrogen bonds present in them. As a result, their boiling point will be lower.
The correct answer is industrial smog. This type of smog exists in coal power plants which creates smoke and sulfur dioxide which may mix with fog creating a thick blanket of haze. Sulfur dioxide is one primary component of an industrial smog.
This because ε = dΦ/dt, and emf is a voltage so it can also be written as dΦ/dt = IR
Because ΔΦ= ABcosθ for each of the N loops and cosθ is just 1 here only A needs to be replaced, and it can be replaced with A=πr^2, which makes that part of the equation ΔΦ_m=N*B0*πa^2
