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tatuchka [14]
4 years ago
6

Identify the type of bonding molecular geometry and intermolecular forced experienced by the compounds hf and hbr which substanc

e would experience a stronger attraction between its molecules and why
Chemistry
1 answer:
Serhud [2]4 years ago
6 0
Both HF and HBr experience H-bond, dipole, and dispersion forces. These forces allow the compounds to have stronger bonds with each other. HF would experience a higher amount of attraction between its molecules. This is because fluorine is more electronegative than bromine.

Hope this helps!
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HELPP
slamgirl [31]

Answer:

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Explanation:

tally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:

\displaystyle \%\text{H}=\frac{\text{mass H}}{\text{mass compound}}\times 100\%%H=

mass compound

mass H

×100%

\displaystyle \%\text{C}=\frac{\text{mass C}}{\text{mass compound}}\times 100\%%C=

mass compound

mass C

×100%

If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:

\displaystyle \%\text{H}=\frac{2.5\text{g H}}{10.0\text{g compound}}\times 100\%=25\%%H=

10.0g compound

2.5g H

×100%=25%

\displaystyle \%\text{C}=\frac{7.5\text{g C}}{10.0\text{g compound}}\times 100\%=75\%%C=

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