Given substances f2, kr, cl2 and hbr rank them with respect to the highest boiling point?

Chemistry

1 answer:

Inessa05 [86]1 year ago

3 0

The Hbr has highest boiling point followed by Kr , $Cl_{2}$ and $F_{2}$ .

The Hbr has highest boiling point due to presence of intermolecular H-bonding that is hydrogen bonding.

The electronegativity of bromine is greater than hydrogen atom. The electronegativity difference hydrogen atom and bromine atom is high. So, due to this a positive dipole is generated at hydrogen atom and partial negative charge, these partial charge is called dipole. In a HBr molecule one H-atom attracted by Br-atom of another atom by force of attraction and this partial force of attraction is called H-bonding.

Other all given molecules are arrange according to its molecular mass because boiling points is directly proportional to the molecular mass of molecule.

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Answer:

Approximately $2.46\; \rm mol$ .

Explanation:

Make use of the molar mass data ( $M({\rm C_2H_2}) = 26.04\; \rm g \cdot mol^{-1}$ ) to calculate the number of moles of molecules in that $64.0\; \rm g$ of $\rm C_2H_2$ :

$\begin{aligned}n({\rm C_2H_2}) &= \frac{m({\rm C_2H_2})}{M} \\ &= \frac{64.0\; \rm g}{26.04\; \rm g\cdot mol^{-1}}\approx 2.46\; \rm mol\end{aligned}$ .

Make sure that the equation for this reaction is balanced.

Coefficient of $\rm C_2H_2$ in this equation: $2$ .

Coefficient of $\rm H_2O$ in this equation: $2$ .

In other words, for every two moles of $\rm C_2H_2$ that this reaction consumes, two moles of $\rm H_2O$ would be produced.

Equivalently, for every mole of $\rm C_2H_2$ that this reaction consumes, one mole of $\rm H_2O$ would be produced.

Hence the ratio: $\displaystyle \frac{n({\rm H_2O})}{n({\rm C_2H_2})} = \frac{2}{2} = 1$ .

Apply this ratio to find the number of moles of $\rm H_2O$ that this reaction would have produced:

$\begin{aligned}n({\rm H_2O}) &= n({\rm C_2H_2}) \cdot \frac{n({\rm H_2O})}{n({\rm C_2H_2})} \\ &\approx 2.46\; \rm mol \times 1 = 2.46\; \rm mol\end{aligned}$ .