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3 years ago

Which of the following molecules would be the best hydrogen bond donor?

Chemistry

1 answer:

leva [86]3 years ago

4 0

Answer: Option (d) is the correct answer.

Explanation:

When a hydrogen atom comes in contact with an electronegative atom then it results in the formation of a chemical bond.

More is the electronegativity of combining atom, more stronger will be the bond with hydrogen atom. As a result, the compound formed will not easily give up hydrogen atom upon dissociation.

Whereas less is the electronegativity of atom combining with hydrogen atom, easily it will donate the hydrogen atom upon dissociation.

Since, out of the given option sulfur (S) atom has low electronegativity as compared to oxygen and nitrogen atom.

Hence, $CH_{3}SH$ will easily donate hydrogen atom.

Thus, we can conclude that $CH_{3}SH$ molecule would be the best hydrogen bond donor.

You might be interested in

If you are given a 0. 31 g piece of sodium metal to react with water, how many moles of hcl would it take to neutralize the sodi

Sphinxa [80]

The moles of HCl to neutralize the sodium hydroxide produced is 0.0135 mole.

Neutralization or neutralization is a chemical response wherein an acid and a base react quantitatively with each other. In a reaction in water, neutralization outcomes in there being no excess of hydrogen or hydroxide ions gift in the answer.

calculation:-

2Na + 2H₂O -----> 2NaOH + H₂

2 mol or 46g of Na produces 80 grams of NaOH

∴ 0.31 g of Na will produce = 80/46 × 0.31

= 0.54 gram of NaOH.

mol of NaOH = 0.54/40

= 0.0135

Since both Hcl and NaOH have the same valance factor,

1 mole NaOH is needed to neutralize 1 mol HCl

∴ 0.0135 mole of NaOH will need = 0.0135 mole of HCl

mass = 0.0135 × 36.5

= 0.493 grams of HCL.

Learn more about neutralizing here:-brainly.com/question/23008798

#SPJ4

4 0

1 year ago

20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

kirill115 [55]

Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
$V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L$ .

Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

$\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}$ .

Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .