Solution
According to ideal gas equation, n = 2 mol, T = 300 K, P = 2.50 x 10 5 Nm - 2 R = 8.314 Nm K - 1 mol - 1 = 19.95 x 10 - 3 m 3 =19.95dm 3..
According to ideal gas equation, n = 2 mol, T = 300 K, P = 2.50 x 10 5 Nm - 2 R = 8.314 Nm K - 1 mol - 1 = 19.95 x 10 - 3 m 3 =19.95dm 3..Solution
According to ideal gas equation, PV = nRT T = 27 + 273 = 300 K, R=82.1 cm 3 atm K - 1 mol - 1 n = 0.0203 mol = 2.03 x 10 - 2 mol. 6. About 200 cm 3 of a gas is confined in a vessel at 20C and 740 mm Hg pressure. How..
According to ideal gas equation, PV = nRT T = 27 + 273 = 300 K, R=82.1 cm 3 atm K - 1 mol - 1 n = 0.0203 mol = 2.03 x 10 - 2 mol. 6. About 200 cm 3 of a gas is confined in a vessel at 20C and 740 mm Hg pressure. How..Purification of Colloidal Sols
Purification of Colloidal Sols - The colloidal solutions prepared by various methods usually contain electrolytes and other soluble substances as impurities. These impurities if not removed can destabilize the sols. Impurities are removed by the following methods: ..
Solution:
Na - 11; 2, 8, 1Number of the valence shell - 3Therefore, sodium is placed in period 3 of the periodic tab..
Solution
Kinetic energy is given as R = 8.314 JK - 1 mol - 1 , T= 273 - 23 = 250 K 9. Calculate the root mean square speed of methane molecules at 2..
Kinetic energy is given as R = 8.314 JK - 1 mol - 1 , T= 273 - 23 = 250 K 9. Calculate the root mean square speed of methane molecules at 2..Solution
The overall energy changes for the formation of MgCl and MgCl 2 in water can be understood by considering the various steps as: H h y d r a t i o n = -735 kJ mol - 1 Net Energy change = - 46 kJ mol - 1 ..
The overall energy changes for the formation of MgCl and MgCl 2 in water can be understood by considering the various steps as: H h y d r a t i o n = -735 kJ mol - 1 Net Energy change = - 46 kJ mol - 1 ..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 10 L. The equilibrium constant of this reaction is given ..
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 10 L. The equilibrium constant of this reaction is given ..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, ..
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, ..Solution
Consider the equation D H r e a c t i o n = 6 D H f (F) + D H f (S) - D H f (SF 6..
Consider the equation D H r e a c t i o n = 6 D H f (F) + D H f (S) - D H f (SF 6..Solution
Data given: 92.20% of silicon of mass 27.98 amu 4.7% of silicon of mass 28.98 amu 3.1 % of silicon of mass 29.97 amu Average atomic weight = 25.80 + 1.36 + 0.93 = 28.09 amu. Average atomic weight of silicon is 28.09 amu.11. One million silver atoms weigh h..
Data given: 92.20% of silicon of mass 27.98 amu 4.7% of silicon of mass 28.98 amu 3.1 % of silicon of mass 29.97 amu Average atomic weight = 25.80 + 1.36 + 0.93 = 28.09 amu. Average atomic weight of silicon is 28.09 amu.11. One million silver atoms weigh h.. Result
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