A student obtained an unknown metal sample that weighed 65.3 g and at a temperature of 99.8oC, he placed it in a calorimeter con

The formula for a monoprotic acid can be represented as HA, and its reaction with a base is shown as follows: HA + NaOH ---> NaA + H₂O. The stoichiometry between the acid and the base is 1:1. At the point of neutralization, the moles of HA equals the moles of the base. The moles of NaOH that reacted can be calculated as 0.100M / 1000 mL/L x 30.0 mL = 0.003 mol. Consequently, the moles of HA that reacted equal 0.003 mol. The mass of the acid is 0.384 g, yielding a molar mass of 0.384 g / 0.003 mol = 128 g/mol.

The correct answer is: c. N2O, N2O4, N2O5.

According to the law of multiple proportions, also referred to as Dalton's Law, when two elements form compounds, the mass ratios of the second element that combine with a specific mass of the first yield small whole number ratios.

1) For NO, the mass ratio m(N): m(O) is 14: 16, simplified to 7: 8.

2) In N₂O, the ratio m(N): m(O) equates to 2·14: 16, which simplifies to 7: 4.

3) For NO₂, the masses yield m(N): m(O) = 14: 2·16, simplifying to 7: 16.

4) In N₂O₅, the ratio is (2·14): (5·16), which simplifies to 7: 20.

5) For NO₄, the mass ratio is m(N): m(O) = 14: (4·16), which simplifies to 7: 32.

6) N₂O₄ gives a ratio of m(N): m(O) as (2·14): (4·16), simplifying to 7: 16.

A) This means m(N): m(O) = 5.6 g: 3.2 g, simplifying results in 1.75: 1, which further translates to m(N): m(O) = 7: 4.

B) Here, m(N): m(O) is 3.5 g: 8.0 g. Dividing reveals a ratio of 1: 2.285, which alters to m(N): m(O) = 7: 16.

C) Lastly, for m(N): m(O) = 1.4 g: 4.0 g, then adjustments yield a ratio of m(N): m(O) = 7: 20.