Solution:
Cl = 35.5 1 mole of a substance = 22.4L 1 Mole of a substance = 1 GMM 1 GMM of Cl 2 = 71 g 71 g of Cl 2 = 22.4 L 7.1 g of Cl 2 = ? 7.1 g of Cl 2 will occupy a volume of 2.24 lite..
Cl = 35.5 1 mole of a substance = 22.4L 1 Mole of a substance = 1 GMM 1 GMM of Cl 2 = 71 g 71 g of Cl 2 = 22.4 L 7.1 g of Cl 2 = ? 7.1 g of Cl 2 will occupy a volume of 2.24 lite..Solution
(i) Fe in Fe 3 O 4 . Let the oxidation number of Fe be 'x'. (ii) S in Na 2 S 4 O 6 . Let the oxidation number of S be x..
(i) Fe in Fe 3 O 4 . Let the oxidation number of Fe be 'x'. (ii) S in Na 2 S 4 O 6 . Let the oxidation number of S be x..Solution
When mixed, the total volume gets doubled and hence the effective concentrations of the ions would be half of the initial concentration, i.e., in solution [Mg 2 + ] =(0.01/2)=0.005 mol/L [C 2 O 4 2 - ] = (0.02/2) = 0.01 mol/L These ions would react to form sparingly soluble ..
When mixed, the total volume gets doubled and hence the effective concentrations of the ions would be half of the initial concentration, i.e., in solution [Mg 2 + ] =(0.01/2)=0.005 mol/L [C 2 O 4 2 - ] = (0.02/2) = 0.01 mol/L These ions would react to form sparingly soluble ..Solution:
As per the equation, 4 vols. : 5 vols. 4 x 22.4 litres : 5 x 22.4 litres 89.6 litres : 112 litres Volume of oxygen required to oxidize 89.6 litres of ammonia = 122 litres Volume of oxygen required to oxidiz..
As per the equation, 4 vols. : 5 vols. 4 x 22.4 litres : 5 x 22.4 litres 89.6 litres : 112 litres Volume of oxygen required to oxidize 89.6 litres of ammonia = 122 litres Volume of oxygen required to oxidiz..Solution
1 mole of any gaseous substance at NTP occupies 22.4 L. 1.12 L of gaseous substance = 0.5 g The molar mass of the substance therefore is 10 g/m..
1 mole of any gaseous substance at NTP occupies 22.4 L. 1.12 L of gaseous substance = 0.5 g The molar mass of the substance therefore is 10 g/m..Solution
As the metal becomes passive by reaction with concentrated HNO 3 it should be iron or component of iron (steel). This metal displaces copper ions from copper (II) sulphate solution, so it is more electropositive, Iron is a grey metal and its oxide Fe 3 O 4 is black. The rea..
As the metal becomes passive by reaction with concentrated HNO 3 it should be iron or component of iron (steel). This metal displaces copper ions from copper (II) sulphate solution, so it is more electropositive, Iron is a grey metal and its oxide Fe 3 O 4 is black. The rea..Solution:
As per the equation, 12 g 1 vol. + 1 vol. 12 g (22.4 + 22.4) litres 12 g 44.8 litres Volume of water gas formed from 12 g of coke = 44.8 litres Volume of water gas formed from 96 kg of coke =? C : Water gas 12 g : 44.8 litres 96000 g : x litres Volume of..
As per the equation, 12 g 1 vol. + 1 vol. 12 g (22.4 + 22.4) litres 12 g 44.8 litres Volume of water gas formed from 12 g of coke = 44.8 litres Volume of water gas formed from 96 kg of coke =? C : Water gas 12 g : 44.8 litres 96000 g : x litres Volume of..Solution:
Mass of 1 mole of Na 2 SO 4 .10H 2 O = (23 x 2) + (32 x 1) + (16 x 4) + 10 (1 x 2 + 16) or, combining the oxygen atoms = 46 + 32 + 64 + 20 + 160 = 322 ..
Mass of 1 mole of Na 2 SO 4 .10H 2 O = (23 x 2) + (32 x 1) + (16 x 4) + 10 (1 x 2 + 16) or, combining the oxygen atoms = 46 + 32 + 64 + 20 + 160 = 322 ..Solution:
Molecular weight of N 2 = 2 x 14 = 28 g28 g of N 2 at STP occupies = 22.4 L 2.8 g of N 2 at STP = ?28 g of N 2 = 22.4 L2.8g of N 2 = ? 2.8 g of N 2 at STP occupies a volume of 2.24..
Molecular weight of N 2 = 2 x 14 = 28 g28 g of N 2 at STP occupies = 22.4 L 2.8 g of N 2 at STP = ?28 g of N 2 = 22.4 L2.8g of N 2 = ? 2.8 g of N 2 at STP occupies a volume of 2.24..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..See what our Users say :
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