The following data are obtained when dinitrogen and dioxygen react together to form different compounds:
Mass of dioxygen | Mass of dinitrogen | |
(i) | 16 g | 14 g |
(ii) | 32 g | 14 g |
(iii) | 32 g | 28 g |
(iv) | 80 g | 28 g |
(a) Which law of chemical combination is obeyed by the above experimental data? Give its statement.
(b) Fill in the blanks in the following conversions:
(i) 1 km = …………………. mm = …………………. pm
(ii) 1 mg = …………………. kg = …………………. ng
(iii) 1 mL = …………………. L = …………………. dm^{3}
(a) If we fix the mass of N_{2} at 28 g, the masses of O_{2} that will combine with the fixed mass of N_{2} are 32 grams, 64 grams, 32 grams and 80 grams.
The mass of O_{2} bear whole no. ratio of 1: 2: 1: 5. Therefore, the given information obeys the law of multiple proportions.
The law of multiple proportions states, “If 2 elements combine to form more than 1 compound, then the masses of one element that combines with the fixed mass of another element are in the ratio of small whole numbers.”
(b) Convert:
(a) 1 km = ____ mm = ____ pm
\( ∴1 km = 10^{ 6 } mm\)
\(∴ 1 km = 10^{ 15 } pm\)
Therefore, 1 km = \(0^{ 6 }mm = 10^{ 15 }pm\)
(b) 1 mg = ____ kg = ____ ng
1 mg =\(10^{ 6 } ng\)
Therefore, 1 mg =\(10^{ -6 }kg = 10^{ 6 } ng\)
(c) 1 mL = ____ L = ____ \(dm^{ 3 }\)
1 mL =\(10^{ -3 } L\)
1 mL =\(10^{ -3 } dm^{ 3 }\)
Therefore, 1 mL = \(10^{ -3 }L = 10^{ -3 }dm^{ 3 }\)
Answered by Pragya Singh | 1 year agoChlorine is prepared in the laboratory by treating manganese dioxide (MnO_{2}) with aqueous hydrochloric acid according to the reaction:
4 HCl (aq) + MnO_{2}(s) → 2H_{2}O (l) + MnCl_{2}(aq) + Cl_{2} (g)
How many grams of HCl react with 5.0 g of manganese dioxide?
Calcium carbonate reacts with aqueous HCl to give CaCl_{2} and CO_{2} according to the reaction, CaCO_{3} (s) + 2 HCl (aq) → CaCl_{2}(aq) + CO_{2} (g) + H_{2}O(l). What mass of CaCO_{3} is required to react completely with 25 mL of 0.75 M HCl?
A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Find:
(i) Empirical formula
(ii) Molar mass of the gas, and
(iii) Molecular formula
Calculate the number of atoms in each of the following
(i) 52 moles of Ar
(ii) 52 u of He
(iii) 52 g of He
Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:
Isotope | Molar mass | Abundance |
\(\, _{ 36 }\textrm{Ar}\) | \( 35.96755 g \; mol^{ -1 }\) | 0.337 % |
\(\, _{ 38 }\textrm{Ar}\) | \( 37.96272 g \; mol^{ -1 }\) | 0.063 % |
\(\, _{ 40 }\textrm{Ar}\) | \( 39.9624g \; mol^{ -1 }\) | 99.600 % |