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3 years ago

Which statement explains why a molecule of CH4 is nonpolar?

Chemistry

2 answers:

Deffense [45]3 years ago

8 0

The answer is the option (3) The geometric shape of a CH4 molecule distributes the charges geometrically. the CH4 is a tetrahedral with the Carbon atom in the center and the four H atoms in the vertices of the tetrahedral. The C-H bond angles are all identical which makes the molecule perfectly symmetrical and so any dipolar moment will cancel making the molecule non polar.

lara [203]3 years ago

5 0

The following statement explains why a molecule of $\text{CH}_4$ is nonpolar:

$\boxed{\left( 3 \right){\text{ The geometric shape of a C}}{{\text{H}}_4}{\text{ molecule distributes the charge symmetrically}}}$ .

Further explanation:

Covalent Bond

This type of bonding takes place when the bonded atoms mutually share electron pairs between them. It is also called the molecule bond. The chemical compounds formed as a result of this bond are called chemical compounds.

The polarity of any bond is primarily governed by two factors; electronegativity difference and symmetry. A bond is said to be polar if the bonded atoms have a considerable molecular bondelectronegativity difference between them. But if there is very small or no electronegativity difference between the bonded atoms, the resulting bond is nonpolar in nature.

Symmetry is another factor responsible for the polarity of the bond. The charge distribution is symmetrical in case symmetry is present in the molecule and as a result, the molecule is said to be nonpolar. But if there is asymmetry in the shape of the molecule, it will be polar in nature.

The electronegativity difference between carbon and hydrogen atoms in $\text{CH}_4$ molecule is very small or negligible so the bonds between the atoms in $\text{CH}_4$ are nonpolar in nature. But the hydrogen atoms are placed in symmetrical positions with respect to the central carbon atom, imparting symmetry to $\text{CH}_4$ molecule and therefore it is nonpolar (For structure, refer to the attached image).

Learn more:

Identification of ionic bonding: brainly.com/question/1603987
Chemical bonds in NaCl: brainly.com/question/5008811

Answer details:

Grade: High School

Chapter: Ionic and covalent compounds

Subject: Chemistry

Keywords: covalent bond, covalent compounds, CH4, carbon, hydrogen, symmetry, electronegativity difference, nonpolar.

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electrons are fermions, they can only access an energy level if it has not already been occupied by another electron. Thanks to this principle, there is a minimum energy level within the atom and it is not possible for an electron to collapse to the nucleus because it cannot exceed this} threshold.

Explanation:

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3 years ago

Calcium Carbonate reacts with dilute hydrochloric acid. The equation for the reaction is shown. CaCo3 + 2Hcl = Cacl2 + H2O + Co2

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Answer:

Approximately $0.224\;\rm L$ , assuming that this reaction took place under standard temperature and pressure, and that $\rm CO_2$ behaves like an ideal gas. Also assume that the reaction went to completion.

Explanation:

The first step is to find out: which species is the limiting reactant?

Assume that $\rm CaCO_3$ is the limiting reactant. How many moles of $\rm CO_2$ would be produced?

Look up the relative atomic mass of $\rm Ca$ , $\rm C$ , and $\rm O$ on a modern periodic table:

$\rm Ca$ : $40.078$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass of $\rm CaCO_3$ :

$\begin{aligned} & M(\rm CaCO_3) \\ &= 40.078 + 12.011 + 3 \times 15.999 \\&= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the number of moles of formula units in $1\; \rm g$ of $\rm CaCO_3$ using its formula mass:

$\begin{aligned}& n(\mathrm{CaCO_3})\\&= \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} \\ &= \frac{1\; \rm g}{100.086\; \rm g \cdot mol^{-1}} \approx 1.00\times 10^{-2}\; \rm mol\end{aligned}$ .

In the balanced chemical equation, the ratio between the coefficient of $\rm CaCO_3$ and that of $\rm CO_2$ is $\displaystyle \frac{n(\mathrm{CO_2})}{n(\mathrm{CaCO_3})} = 1$ .

In other words, for each mole of $\rm CaCO_3$ formula units consumed, one mole of $\rm CO_2$ would be produced.

If $\rm CaCO_3$ is indeed the limiting reactant, all that approximately $1.00\times 10^{-2}\; \rm mol$ of $\rm CaCO_3\!$ formula would be consumed. That would produce approximately $1.00\times 10^{-2}\; \rm mol\!$ of $\rm CO_2$ .

On the other hand, assume that $\rm HCl$ is the limiting reactant.

Convert the volume of $\rm HCl$ to $\rm dm^{3}$ (so as to match the unit of concentration.)

$\begin{aligned}&V(\mathrm{HCl})\\ &= 50\; \rm cm^{3} \\ &= 50\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{10^{3}\; \rm cm^{3}} \\ &= 5.00\times 10^{-2}\; \rm dm^{3} \end{aligned}$ .

Calculate the number of moles of $\rm HCl$ molecules in that $5.00\times 10^{-2}\; \rm dm^{3}$ of this $\rm 0.05\; \rm mol \cdot dm^{-3}$

$\begin{aligned}& n(\mathrm{HCl}) \\ &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.05\; \rm mol \cdot dm^{-3}\\ &\quad\quad \times 5.00\times 10^{-2}\;\rm dm^{3} \\ &= 2.50 \times 10^{-3}\; \rm mol\end{aligned}$ .

Notice that in the balanced chemical reaction, the ratio between the coefficient of $\rm HCl$ and that of $\rm CO_2$ is $\displaystyle \frac{n(\mathrm{CO_2})}{n(\mathrm{HCl})} = \frac{1}{2}$ .

In other words, each mole of $\rm HCl$ molecules consumed would produce only $0.5\;\rm mol$ of $\rm CO_2$ molecules.

Therefore, if $\rm HCl$ is the limiting reactant, that $2.50 \times 10^{-3}\; \rm mol$ of $\rm HCl\!$ molecules would produce only one-half as many (that is, $1.25\times 10^{-3}\; \rm mol$ ) of $\rm CO_2$ molecules.

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In reality, no more than $\rm 1.00\times 10^{-3}\; \rm mol$ of $\rm CO_2$ molecules would be produced. The reason is that all $\rm CaCO_3$ would have been consumed before $\rm HCl$ was.

After finding the limiting reactant, approximate the volume of the $\rm CO_2\!$ produced.

Assume that this reaction took place under standard temperature and pressure (STP.) Under STP, the volume of one mole of ideal gas molecules would be approximately $22.4\; \rm L$ .

If $\rm CO_2$ behaves like an ideal gas, the volume of that $\rm 1.00\times 10^{-3}\; \rm mol$ of $\rm CO_2\!$ molecules would be approximately $\rm 1.00\times 10^{-3}\; \rm mol \times 22.4\; \rm L = 0.224\; \rm L$ .

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