<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
<span>Equation:2H2(g) + O2(g) → 2H2O(g)
</span><span>
Smaller container means less volume, and the molecules will hit the walls of the container more frequently because there's less space available and the pressure will go up. I guess this would mean that the side with fewer moles would be favored as a result. We count the number of moles on the reactants and products and find that there are fewer moles on the product side, so I guess this would favor the product formation.
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Molar mass of N = 14 g/molMolar mass of O2 = 32 g/molAdding both masses = 46 g/molActual molar mass/ Empirical molar mass = 138.02 / 46 = 3Now multiplying this co effecient with empirical fomula NO2 = 3(NO2) = N3O6So according to above explanation,D) N3O6, is the correct answer.
<span>Metals tend to lose electrons and form electro-positive ions / cations.</span>
Since
21.2 g H2O was produced, the amount of oxygen that reacted can be obtained
using stoichiometry. The balanced equation was given: 2H₂ + O₂ → 2H₂O and
the molar masses of the relevant species are also listed below. Thus, the
following equation is used to determine the amount of oxygen consumed.
Molar mass of H2O = 18
g/mol
Molar mass of O2 = 32
g/mol
21.2 g H20 x 1 mol
H2O/ 18 g H2O x 1 mol O2/ 2 mol H2O x 32 g O2/ 1 mol O2 = 18.8444 g O2
<span>We then determine that
18.84 g of O2 reacted to form 21.2 g H2O based on stoichiometry. It is
important to note that we do not need to consider the amount of H2 since we can
derive the amount of O2 from the product. Additionally, the amount of H2 is in
excess in the reaction.</span>