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Solution
The reaction is, Initial amount: 1 mol 1 mol = 0.4 mol = 0.4 mol Amounts at equilibrium:(1-0.4)mol (1-0.4)mol 0.4 mol 0.4 mol = 0.6 mol = 0.6 mol 0.4 mol 0.4 mol Volume of the reaction vessel is..
The reaction is, Initial amount: 1 mol 1 mol = 0.4 mol = 0.4 mol Amounts at equilibrium:(1-0.4)mol (1-0.4)mol 0.4 mol 0.4 mol = 0.6 mol = 0.6 mol 0.4 mol 0.4 mol Volume of the reaction vessel is..Solution:
1 GMM of water (H 2 O) = 2 x 1 + 16 = 18 g.18 g = 1 moleXg = 0.4 moles 0.4 moles of water weigh 7.2..
1 GMM of water (H 2 O) = 2 x 1 + 16 = 18 g.18 g = 1 moleXg = 0.4 moles 0.4 moles of water weigh 7.2..Solution
(I) Sulphur atom has six valence electrons. The two hydrogens in this compound are present as OH groups. The possible Lewis structure of H 2 SO 4 is, (III) Boron atom (B) has 3 electrons in its outermost shell, and Cl has 7. B does not obey the octet rule. The possible Lewis structu..
(I) Sulphur atom has six valence electrons. The two hydrogens in this compound are present as OH groups. The possible Lewis structure of H 2 SO 4 is, (III) Boron atom (B) has 3 electrons in its outermost shell, and Cl has 7. B does not obey the octet rule. The possible Lewis structu..Solution:
As per the equation, 2 (14 + 4 + 35.5) 2 vols.(of NH 3 ) 107 2 x 22.4 litres 107 g 44.8 litres or 44800 ml NH 4 Cl : NH 3 107 g : 44800 ml 2.675 g : x Volume of ammonia formed at STP = 1120 ml. Example: 31Calculate the volume of sulp..
As per the equation, 2 (14 + 4 + 35.5) 2 vols.(of NH 3 ) 107 2 x 22.4 litres 107 g 44.8 litres or 44800 ml NH 4 Cl : NH 3 107 g : 44800 ml 2.675 g : x Volume of ammonia formed at STP = 1120 ml. Example: 31Calculate the volume of sulp..Solution
Liquation is a method for concentrating ores, which have a lower melting point than the impurities. The ores of antimony are concentrated by this method. The powdered ore is heated upon a sloping floor of the furnace. The temperature is raised above the melting point of the ore: this causes the or..
Solution:
1 GMM of water (H 2 O) = 2 x 1 + 16 = 18g. 18g = 1 mole Xg = 0.4 moles 0.4 moles of water weighs 7.2..
1 GMM of water (H 2 O) = 2 x 1 + 16 = 18g. 18g = 1 mole Xg = 0.4 moles 0.4 moles of water weighs 7.2..Solution:
The given Problem is solved in the following table. Empirical formula = N 2 H 4 O 3 Empirical formula mass = (14 x 2) + (1 x 4) + (16 x 3) = 28 + 4 + 48 = 80 Molecular mass = 80 Ratio of Molecular mass to Empirical formula mass \ ..
The given Problem is solved in the following table. Empirical formula = N 2 H 4 O 3 Empirical formula mass = (14 x 2) + (1 x 4) + (16 x 3) = 28 + 4 + 48 = 80 Molecular mass = 80 Ratio of Molecular mass to Empirical formula mass \ ..Solution
In the resulting solution, [Ca + + ] = 0.0025M. [C 2 O 4 2 - ] = 1 x 10 - 7 M. [Ca + + ][C 2 O 4 2 - ] = 0.0025 x 10 - 7 = 2.5 x 10 - 1 0 This is less than the solubility product of calcium oxalate. Thus, precipitation of ca..
Solution
The degree of hardness of water is defined as the number of parts by mass of calcium carbonate, equivalent to various calcium and magnesium present in one million parts of mass of water. It is expressed in ppm. Now, 10 6 g of water contains CaCO 3 = 44 mg By definition, 1 mol of CaCO 3..
The degree of hardness of water is defined as the number of parts by mass of calcium carbonate, equivalent to various calcium and magnesium present in one million parts of mass of water. It is expressed in ppm. Now, 10 6 g of water contains CaCO 3 = 44 mg By definition, 1 mol of CaCO 3..Solution:
Element % by wt At wt Relative No.of Atoms Simple Ratio Fe 72.41 56 72.41/56 = 1.29 1.29/1.29 = 1 x 3 = 3 O 27.59 16 27.59/16 = 1.72 1.72/1.29 = 1.33 x 3 = 4 Therefore simple ratio = Fe 3 O 4 .Empirical formula = Fe 3 O 4..
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