The solutions of the quadratic equation x2 + px + q = 0 are l and m. W..
The solutions of the quadratic equation x 2 + p x + q = 0 are l and m . Write a quadratic equation that has 1 l and 1 m as the solutions. => p x 2 + q x + 1 = 0 or q x 2 + p q x + 1..
Solution
(a) 38% solution, means 38 g of HCl in 100 g of solution. Then, Mass of the solution= 100 g Molar mass of HCl = 36.5 g mol - 1 (b) The molarity of conc. HCl sample = 12.38 mol/L Molarity of HCl solution to be prepared = 0.10 mol/h..
(a) 38% solution, means 38 g of HCl in 100 g of solution. Then, Mass of the solution= 100 g Molar mass of HCl = 36.5 g mol - 1 (b) The molarity of conc. HCl sample = 12.38 mol/L Molarity of HCl solution to be prepared = 0.10 mol/h..Solution
2 M solution of H 2 O 2 means 2 mol of H 2 O 2 in one litre of the solution. Molar mass of H 2 O 2 = (2x1 + 2x16) g/mol =(2 + 32) g/mol = 34 g/mol Thus, there are 68 g (= 2 x 34 g) of H 2 O 2 in one litre of the solution. So..
2 M solution of H 2 O 2 means 2 mol of H 2 O 2 in one litre of the solution. Molar mass of H 2 O 2 = (2x1 + 2x16) g/mol =(2 + 32) g/mol = 34 g/mol Thus, there are 68 g (= 2 x 34 g) of H 2 O 2 in one litre of the solution. So..Solution
P 1 = 745 mm Hg V 1 = 250 mL P 2 = ? V 2 = 200mL P 1 V 1 = P 2 V 2 The additional pressure required is 931.25 - 745 = 186.25 ..
P 1 = 745 mm Hg V 1 = 250 mL P 2 = ? V 2 = 200mL P 1 V 1 = P 2 V 2 The additional pressure required is 931.25 - 745 = 186.25 ..Solution
(i) HCl completely ionizes as: The concentration of hydronium ions is equal to that of hydrochloric acid, [H 3 O + ] = [HCl] [HCl] = 0.01M = 1 x 10 - 2 mol L - 1 The ionic product of water is K w = [H 3 O + ] [OH - ] Thus [H 3 O + ] = 1 x ..
(i) HCl completely ionizes as: The concentration of hydronium ions is equal to that of hydrochloric acid, [H 3 O + ] = [HCl] [HCl] = 0.01M = 1 x 10 - 2 mol L - 1 The ionic product of water is K w = [H 3 O + ] [OH - ] Thus [H 3 O + ] = 1 x ..Solution
The volume of H 2 SO 4 neutralised by N/10 NaOH can be obtained as follows: N a c i d x V a c i d = N a l k a l i x V a l k a l i Therefore, Volume of N/5 H 2 SO 4 used for neutralizing NH 3 = (100 - 77) mL = 23 mLh..
The volume of H 2 SO 4 neutralised by N/10 NaOH can be obtained as follows: N a c i d x V a c i d = N a l k a l i x V a l k a l i Therefore, Volume of N/5 H 2 SO 4 used for neutralizing NH 3 = (100 - 77) mL = 23 mLh..Solution
The second electron in the case of the alkali metals (Na) is to be removed from a cation, which has already acquired a noble gas configuration. On the other hand, in the alkaline Earth metal (Mg), the second electron is to be removed from a monovalent cation (Mg + ), which still has one..
The second electron in the case of the alkali metals (Na) is to be removed from a cation, which has already acquired a noble gas configuration. On the other hand, in the alkaline Earth metal (Mg), the second electron is to be removed from a monovalent cation (Mg + ), which still has one..Solution
D H c o m b of Glucose (C 6 H 1 2 O 6 ) = - 2880 kJ mol - 1 14. Calculate the enthalpy change of combustion of cyclopropane at 298 K. The enthalpy of formation of CO 2 ( g ) , H 2 O ( l ) and propane ( g ) are -393.5, -285.8 and 20.42..
D H c o m b of Glucose (C 6 H 1 2 O 6 ) = - 2880 kJ mol - 1 14. Calculate the enthalpy change of combustion of cyclopropane at 298 K. The enthalpy of formation of CO 2 ( g ) , H 2 O ( l ) and propane ( g ) are -393.5, -285.8 and 20.42..Choose a quadratic equation that has the solutions 13 and 9.
Choose a quadratic equation that has the solutions 13 and 9. => x 2 - 22 x + 4 = 0 or x 2 - 4 x + 117 = 0 or x 2 - 22 x + 117 = 0 or x 2 + 22 x + 117 = 0..
Equivalent Conductance (^eq.)
Equivalent Conductance (^eq.) - It is defined as the conductance of that volume of solution which has one equivalent dissolved in it. V = Volume of solution which has one equivalent in it. N normality (Number of equivalents per liter or 1000c..
Equivalent Conductance (^eq.) - It is defined as the conductance of that volume of solution which has one equivalent dissolved in it. V = Volume of solution which has one equivalent in it. N normality (Number of equivalents per liter or 1000c..See what our Users say :
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