Take a hypothetical sample of exactly 100 grams of the solution.
(16g urea) / (60.06 g urea/mol) = 0.2664 mol urea
((100 g total) - (16g urea)) = 84.0 g H2O = 0.0840 kg H2O
(0.2664 mol) /0.0840 (kg) = 3.17143mol/kg = 3.18m urea
This is a combustion reaction.
2C6H6 + 9O2➡️ 6CO2 + 6H2O
BF3 .... BP = −100.3 °C
<span>RbCl ..... solid </span>
<span>CH3SCH3 ..... BP = 35-41 °C </span>
<span>SbH3 .... BP = −17 °C </span>
<span>SiS2 ..... solid </span>
<span>Ethanol solid --> ethanol melts --> ethanol liquid </span>
<span>-135C ---------------> -114C --------------> -50C </span>
<span>............ ΔT = 21C ....... ....... ΔT = 64C </span>
Answer:
The three compounds are different compounds
Explanation:
The mass of Nitrogen that combines with 1 gram of Oxygen in Compound A = 1.750 g
The mass of Nitrogen that combines with 1 gram of Oxygen in Compound B = 0.8750 g
The mass of Nitrogen that combines with 1 gram of Oxygen in Compound C = 0.4375 g
According to the law of multiple proportions, when atoms of two different elements react to form compounds, the masses of one of the elements that combines with a fixed mass of the other element are in small whole number ratios.
The ratio of the masses are;
Mass of Nitrogen in Compound B/(Mass of Nitrogen in Compound C = 0.8750/0.4375 = 2
Mass of Nitrogen in Compound A/(Mass of Nitrogen in Compound C = 1.750/0.4375= 4
Mass of Nitrogen in Compound A/(Mass of Nitrogen in Compound B = 1.750/0.8750= 2
Given that the masses of Nitrogen in the three compounds are in small whole number ratios, the three compounds, Compound A, Compound B, and Compound C are different compounds.
