Solution :
(a) Valence electrons = 3; core electrons = 10(b) Valence electrons = 2; core electrons = 10(c) Valence electrons = 7; core electrons = 10(d) Valence electrons = 2; core electrons =..
Solution:
The given problem is solved in the following table. Empirical formula = C 1 H 1 = ..
The given problem is solved in the following table. Empirical formula = C 1 H 1 = ..Solution:
The given problem is solved in the following table. Empirical formula = Ca 1 C 1 O 3 = CaCO..
The given problem is solved in the following table. Empirical formula = Ca 1 C 1 O 3 = CaCO..Solution:
2 vol. : 1vol : 2 vol. 30 litres: x litres : y litres CO : O 2 CO : CO 2 2 : 1 2 : 2 30 : x 30 : y 2x = 30 x 1 2y = 30 x 2 x = 15 y= 30 Volume of oxygen = 15 litres Volume of carbon dioxide = 30 litre..
2 vol. : 1vol : 2 vol. 30 litres: x litres : y litres CO : O 2 CO : CO 2 2 : 1 2 : 2 30 : x 30 : y 2x = 30 x 1 2y = 30 x 2 x = 15 y= 30 Volume of oxygen = 15 litres Volume of carbon dioxide = 30 litre..Solution:
The given problem is solved in the following table. Empirical formula = N 2 H 8 S 1 O 4 = N 2 H 8 SO 4..
The given problem is solved in the following table. Empirical formula = N 2 H 8 S 1 O 4 = N 2 H 8 SO 4..Solution:
Mass of one mole of pure calcium carbonate (CaCO 3 ) = Mass of calcium + Mass of carbon + Mass of oxygen = (40 x 1) + (12 x 1) + (16 x 3) = 40 + 12 + 48 = 100 g ..
Mass of one mole of pure calcium carbonate (CaCO 3 ) = Mass of calcium + Mass of carbon + Mass of oxygen = (40 x 1) + (12 x 1) + (16 x 3) = 40 + 12 + 48 = 100 g ..Solution:
The given problem is solved in the following table. Empirical formula = MgSO 1 1 H 1 4 Empirical formula mass = (24 x 1) + (32 x 1) + (16 x 11) + (1 x 14) = 24 + 32 + 176 + 14 = 246 Molecular mass = 246 Hence, Molecular form..
The given problem is solved in the following table. Empirical formula = MgSO 1 1 H 1 4 Empirical formula mass = (24 x 1) + (32 x 1) + (16 x 11) + (1 x 14) = 24 + 32 + 176 + 14 = 246 Molecular mass = 246 Hence, Molecular form..Solution
1 cm 3 of ethanol weighs 0.79 g or 0.79 x 10 - 3 kg 1 m 3 i.e., 10 6 cm 3 weighs = 0.79 x 10 - 3 x 10 6 kg = 0.79 x 10 3 kg 0.79 g cm - 3 = 0.79 x 10 3 kg m - ..
Solution:
Mass of the reactants = 10gMass of the products = 4.4 + 5.6g = 10gSince the mass of the reactants is equal to the mass of the products, the observations are in agreement with the law of conservation of mas..
Solution:
Gram molecular mass of NaOH = (23 + 16 + 1) = 40 g Mass of 1 mole of NaOH = 40g No. of moles present in 3200 g of NaOH = 80 mol..
Gram molecular mass of NaOH = (23 + 16 + 1) = 40 g Mass of 1 mole of NaOH = 40g No. of moles present in 3200 g of NaOH = 80 mol..See what our Users say :
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