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2 years ago

Why are the molecules of hydrogen chloride gas (hcl) more tightly bound to each other than molecules of oxygen gas (o2)?

1 answer:

4 0

Answer : If we consider the molecule of oxygen gas which in diatomic state, is bonded to other atom which is of same element this is called as homonuclear.

While in HCl there is a heteronuclear bonding observed because there are two different elements getting involved in the bond formation, also it creates a electrovalent species in itself and makes it more polar. They are creating a dipole moment by separating different charges in the molecule which cause it to get tightly bonded with each other.

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25.0cm3 of s saturated potassium hydroxide is neutralized by 35.0cm3 of hydrogen chloride acid of concentration 0.75 mol/dm3. Ca

Kamila [148]

Answer:

Concentration of the original $\rm KOH$ solution: approximately $1.05\; \rm mol \cdot dm^{-3}$ .

Explanation:

Notice that the concentration of the $\rm HCl$ solution is in the unit $\rm mol\cdot dm^{-3}$ . However, the unit of the two volumes is $\rm cm^{3}$ . Convert the unit of the two volumes to $\rm dm^{3}$ to match the unit of concentration.

$\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}$ .

$\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}$ .

Calculate the number of moles of $\rm HCl$ molecules in that $0.0350\; \rm dm^{3}$ of $0.75\; \rm mol \cdot dm^{3}$ $\rm HCl\!$ solution:

$\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}$ .

$\rm HCl$ is a monoprotic acid. In other words, each $\rm HCl\!$ would release up to one proton $\rm H^{+}$ .

On the other hand, $\rm KOH$ is a monoprotic base. Each $\rm KOH\!$ formula unit would react with up to one $\rm H^{+}$ .

Hence, $\rm HCl$ molecules and $\rm KOH\!$ formula units would react at a one-to-one ratio.

${\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l)$ .

Therefore, that $0.02625\; \rm mol$ of $\rm HCl$ molecules would neutralize exactly the same number of $\rm NaOH$ formula units. That is: $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ .

Calculate the concentration of a $\rm NaOH$ solution where $V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3}$ and $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ :

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}$ .

7 0

2 years ago

The sulfur atom of sulfur dioxide is considered to be sp2-hybridized. The expected bond angle is 120°, but is actually slightly

Alja [10]

Answer: The bonds are intermediate between double and single bonds

Explanation:

A closer look at the diagram below shows that the bonds in sulphur IV oxide are intermediate between double and single bonds. Hence they do not have the exact bond angle of single bonds. This is why the bond angle is not exactly 120°. There are two resonance structures in the diagram that clearly show this point.

6 0

2 years ago

What are the covalent bonds called that form between atoms of different electronegativities?

ZanzabumX [31]

Polar bonds are formed between atoms of different electronegativities.

5 0

2 years ago

A sample of a gas is contained in a closed rigid cylinder. According to kinetic molecular theory, what occurs when the gas insid

Anna11 [10]

The answer is (3) The average velocity of the gas molecules increases. In the closed rigid cylinder, the volume of the gas and number of gas molecules will not change. The number of collision will increase.

4 0

3 years ago

Read 2 more answers

Bromine has two naturally occurring isotopes (br-79 and br-81) and has an atomic mass of

Rzqust [24]

Answer: 78.919 amu

Explanation:

If the natural abundance of Br-81 is 49.310%, then the natural abundance of Br-79 must be 100%-49.310%=50.690%

Substituting into the atomic mass formula,

$79.904=(0.49310)(80.9163)+(0.50690)(x)\\79.904=39.89982753+(0.50690)(x)\\40.00417247=(0.50690)(x)\\x=\frac{40.00417247}{0.50690}=\boxed{78.919 \text{ amu}}$

5 0

1 year ago