Answer:
At high temperatures or in the presence of catalysts, sulfur dioxide reacts with hydrogen sulfide to form elemental sulfur and water. This reaction is exploited in the Claus process, an important industrial method to dispose of hydrogen sulfide.
<u>Answer:</u>
(A)
Density = Mass / Volume
So
Mass = Density × Volume
![= 0.867 g/mL \times 43.3mL = 37.5411 g Toluene](https://tex.z-dn.net/?f=%3D%200.867%20g%2FmL%20%5Ctimes%2043.3mL%20%3D%2037.5411%20g%20Toluene)
![1C_6 H_5 CH_3 + 9 O_2 > 7 CO_2 + 4 H_2 O](https://tex.z-dn.net/?f=1C_6%20H_5%20CH_3%20%20%2B%209%20O_2%20%20%3E%207%20CO_2%20%20%2B%204%20H_2%20O)
Mole ratio of toluene : Oxygen is 1 : 9
![$37.5411 g \text { Toluene } \times \frac{1 \text {mol} \text {toluene}}{92 g \text { toluene}} \times \frac{9 {mol} O_{2}}{1 \text {mol} \text { toluene }} \times \frac{32 g O_{2}}{1 {mol} O_{2}}=117 g O_{2}(\text {Answer})$](https://tex.z-dn.net/?f=%2437.5411%20g%20%5Ctext%20%7B%20Toluene%20%7D%20%5Ctimes%20%5Cfrac%7B1%20%5Ctext%20%7Bmol%7D%20%5Ctext%20%7Btoluene%7D%7D%7B92%20g%20%5Ctext%20%7B%20toluene%7D%7D%20%5Ctimes%20%5Cfrac%7B9%20%7Bmol%7D%20O_%7B2%7D%7D%7B1%20%5Ctext%20%7Bmol%7D%20%5Ctext%20%7B%20toluene%20%7D%7D%20%5Ctimes%20%5Cfrac%7B32%20g%20O_%7B2%7D%7D%7B1%20%7Bmol%7D%20O_%7B2%7D%7D%3D117%20g%20O_%7B2%7D%28%5Ctext%20%7BAnswer%7D%29%24)
(B)
1 mole of Toluene produces 7 moles of
gas and 4 moles of
Vapour
So the mole ratio is 1 : 11
![37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene }} \times \frac{11 \mathrm{mol} \text { gas }}{1 \text { mol toluene }} $$\\\\=4.49 \text { mol gaseous products (Answer) } $](https://tex.z-dn.net/?f=37.5411%20g%20Toluene%20%24%5Ctimes%20%5Cfrac%7B1%20%5Ctext%20%7B%20mol%20toluene%20%7D%7D%7B92%20g%20%5Ctext%20%7B%20toluene%20%7D%7D%20%5Ctimes%20%5Cfrac%7B11%20%5Cmathrm%7Bmol%7D%20%5Ctext%20%7B%20gas%20%7D%7D%7B1%20%5Ctext%20%7B%20mol%20toluene%20%7D%7D%20%24%24%5C%5C%5C%5C%3D4.49%20%5Ctext%20%7B%20mol%20gaseous%20products%20%28Answer%29%20%7D%20%24)
(C)
1mole contains
molecules
![37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene}} \times \frac{4 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \text { toluene }} \times \frac{6.022 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}} $\\\\$=9.82 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O} \text { (Answer) } $](https://tex.z-dn.net/?f=37.5411%20g%20Toluene%20%24%5Ctimes%20%5Cfrac%7B1%20%5Ctext%20%7B%20mol%20toluene%20%7D%7D%7B92%20g%20%5Ctext%20%7B%20toluene%7D%7D%20%5Ctimes%20%5Cfrac%7B4%20%5Cmathrm%7Bmol%7D%20%5Cmathrm%7BH%7D_%7B2%7D%20%5Cmathrm%7BO%7D%7D%7B1%20%5Cmathrm%7Bmol%7D%20%5Ctext%20%7B%20toluene%20%7D%7D%20%5Ctimes%20%5Cfrac%7B6.022%20%5Ctimes%2010%5E%7B23%7D%20%5Ctext%20%7B%20molecules%20%7D%20%5Cmathrm%7BH%7D_%7B2%7D%20%5Cmathrm%7BO%7D%7D%7B1%20%5Cmathrm%7Bmol%7D%20%5Cmathrm%7BH%7D_%7B2%7D%20%5Cmathrm%7BO%7D%7D%20%24%5C%5C%5C%5C%24%3D9.82%20%5Ctimes%2010%5E%7B23%7D%20%5Ctext%20%7B%20molecules%20%7D%20%5Cmathrm%7BH%7D_%7B2%7D%20%5Cmathrm%7BO%7D%20%5Ctext%20%7B%20%28Answer%29%20%7D%20%24)
Yes, o-toluic acid is soluble in ether as ether is slightly polar and it is soluble in NaOH because it is likely to form soluble compounds with it.
Naphthalene is insoluble in NaOH.
Answer:
They are multicelled
Explanation:
I just did a quiz on it UwU