Answer:
![\boxed{\text{(3) 9.6 L}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Ctext%7B%283%29%209.6%20L%7D%7D)
Explanation:
1. Moles of CCl₄
![n = 6.02 \times 10^{25} \text{ molecules} \times \dfrac{\text{1 mol}}{6.022 \times 10^{23}\text{ molecules}} = \text{100.0 mol}](https://tex.z-dn.net/?f=n%20%3D%206.02%20%5Ctimes%2010%5E%7B25%7D%20%5Ctext%7B%20molecules%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%7D%7D%7B6.022%20%5Ctimes%2010%5E%7B23%7D%5Ctext%7B%20molecules%7D%7D%20%3D%20%5Ctext%7B100.0%20mol%7D)
2. Molar mass of CCl₄
MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol
3. Mass of CCl₄
![m =\text{100.0 mol} \times \dfrac{\text{154.0 g}}{\text{1 mol}} = \text{15 400 g}](https://tex.z-dn.net/?f=m%20%3D%5Ctext%7B100.0%20mol%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B154.0%20g%7D%7D%7B%5Ctext%7B1%20mol%7D%7D%20%3D%20%5Ctext%7B15%20400%20g%7D)
4. Volume of CCl₄
![V = \text{15 400 g} \times \dfrac{\text{1 cm}^{3}}{\text{1.6 g}} = \text{9600 cm}^{3}\\\\V = \text{9600 cm}^{3} \times \dfrac{\text{1 L}}{\text{1000 cm}^{3}} = \mathbf{{9.6 L}}\\\\\text{The volume of CCl$_{4}$ is } \boxed{\textbf{9.6 L}}](https://tex.z-dn.net/?f=V%20%3D%20%5Ctext%7B15%20400%20g%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20cm%7D%5E%7B3%7D%7D%7B%5Ctext%7B1.6%20g%7D%7D%20%3D%20%5Ctext%7B9600%20cm%7D%5E%7B3%7D%5C%5C%5C%5CV%20%3D%20%5Ctext%7B9600%20cm%7D%5E%7B3%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20L%7D%7D%7B%5Ctext%7B1000%20cm%7D%5E%7B3%7D%7D%20%3D%20%5Cmathbf%7B%7B9.6%20L%7D%7D%5C%5C%5C%5C%5Ctext%7BThe%20volume%20of%20CCl%24_%7B4%7D%24%20is%20%7D%20%5Cboxed%7B%5Ctextbf%7B9.6%20L%7D%7D)
Polar molecules exhibit an unequal balance of charges between the individual elements of the compound. This is brought about by the large difference in their electronegativities. The H atom has the least amount of electronegativity. Then, it is a known periodic trend, that as you go downwards in a group, electronegativity decreases, and increase as you go from left to right. Thus, you can deduce that the most electronegative elements are found in the upper right corner which includes O, N and F atoms. Any bond created between Hydrogen and any of O, N and F atoms is a polar bond.
<u>Answer:</u> The expression for equilibrium constant is ![K_{eq}=\frac{[HOCl]^2}{[H_2O][Cl_2]^2}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BHOCl%5D%5E2%7D%7B%5BH_2O%5D%5BCl_2%5D%5E2%7D)
<u>Explanation:</u>
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as ![K_{eq}](https://tex.z-dn.net/?f=K_%7Beq%7D)
For the general chemical equation:
![aA+bB\rightleftharpoons cC+dD](https://tex.z-dn.net/?f=aA%2BbB%5Crightleftharpoons%20cC%2BdD)
The expression for
is given as:
![K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
For the given chemical reaction:
![2HgO(s)+H_2O(l)+2Cl_2(g)\rightleftharpoons 2HOCl(aq.)+HgO.HgCl_2(s)](https://tex.z-dn.net/?f=2HgO%28s%29%2BH_2O%28l%29%2B2Cl_2%28g%29%5Crightleftharpoons%202HOCl%28aq.%29%2BHgO.HgCl_2%28s%29)
The expression for
is given as:
![K_{eq}=\frac{[HOCl]^2[HgO.HgCl_2]}{[HgO]^2[H_2O][Cl_2]^2}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BHOCl%5D%5E2%5BHgO.HgCl_2%5D%7D%7B%5BHgO%5D%5E2%5BH_2O%5D%5BCl_2%5D%5E2%7D)
The concentration of solid is taken to be 0.
So, the expression for
is given as:
![K_{eq}=\frac{[HOCl]^2}{[H_2O][Cl_2]^2}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BHOCl%5D%5E2%7D%7B%5BH_2O%5D%5BCl_2%5D%5E2%7D)
Answer:
<u>a</u><u>.</u><u> </u><u>True</u><u>.</u>
Explanation:
Only primary and secondary alcohols can oxidise to give an aldehyde. But a weak oxidizing agent must be used to prevent formation of a carboxylic acid or ketone.
weak oxidizing agents: Chromyl chloride, silver/oxygen/500°C
take an example of <u>e</u><u>t</u><u>h</u><u>a</u><u>n</u><u>o</u><u>l</u><u>:</u>
<u>
</u>
<u>
</u>
<u>B</u><u>y</u><u> </u><u>o</u><u>z</u><u>o</u><u>n</u><u>o</u><u>l</u><u>y</u><u>s</u><u>i</u><u>s</u><u>:</u>
Here, reactants are Ozone gas, Carbon tetrachloride at a temperature (<20°C), ethanoic acid, zinc and water.
take an example of propanol:
if it undergoes ozonolysis, it gives ethanal and methanal.