<u>Answer:</u> The freezing point of solution is 2.6°C
<u>Explanation:</u>
To calculate the depression in freezing point, we use the equation:
Or,
where,
= 
Freezing point of pure solution = 5.5°C
i = Vant hoff factor = 1 (For non-electrolytes)
= molal freezing point depression constant = 5.12 K/m = 5.12 °C/m
= Given mass of solute (anthracene) = 7.99 g
= Molar mass of solute (anthracene) = 178.23 g/mol
= Mass of solvent (benzene) = 79 g
Putting values in above equation, we get:
Hence, the freezing point of solution is 2.6°C
D) Contain Chemical bonds.
Answer:
2 C4H10 + 5 O2 → 4 CH3CO2H + 2 H2O.
Explanation:
Light naphtha components are readily oxidized by oxygen or even air to give peroxides, which decompose to produce acetic acid according to the chemical equation, illustrated with butane .
ionic bond is formed between ca and cl forming molecule cacl2 ca has 2 velancy and cl has one velancy (ca has 2 electrons in its outer most shell while cl has 1 electron vecancy in its outermost shell). So ca would make bond with 2 cl atoms
Monoprotic acid are acids having
only one hydrogen atoms after dissociation into ions from its compound. The
monoprotic acid from among the following is HCl. The answer is letter D. HCl →
H+ + Cl-. Note that there is only one H+ ion upon dissociation.
