the solutions of the quadratic equation . write a quadratic equation that has l 2 and m 2 as the solutions.

The solutions of the quadratic equation x 2 + p x + q = 0 are l and m . Write a quadratic equation that has l + 2 and m + 2 as the solutions. => x 2 +2 p x + 4 q = 0 or ( x - ..

Identify a quadratic equation with solutions 2 and - 4. => x 2 - 2 x + 8 = 0 or x 2 + 2 x + 8 = 0 or x 2 - 2 x - 8 = 0 or x 2 + 2 x - 8 = 0..

The solutions of the quadratic equation x 2 + p x + q = 0 are l and m . Write a quadratic equation that has - l and - m as the solutions. => x 2 - p x - q = 0 or x 2 + p x - q = 0 or x 2 - ..

(a) For n = 3, the permissible values for 'l' and 'm' are: l = 0, 1, 2 For 'l' = 0 m = 0 (s-orbit..

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, th..

200 mL of CO 2 = 0.40 g 22.4 L of CO 2 = 12 + 32 (44g) Molecular weight = 2 x vapor density = 2 x 22.22 = 44..

Electron configuration: K L M 2 8 7 Number of valence electrons = 7Valency = 8 - 7 = 1Valence shell = M..

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) m..

Mass of the substance taken = 0.2 g Volume of air displaced = 30 mL Temperature = 27 o C = (27 + 273) K = 300 K Atmospheric pressure = (756 - 26) mm Hg = 730 mm Hg Then, Therefore, Volume of the vapours at NTP = 22.5 mL Then, ..

Mass of the organic compound taken = 0.244 g Mass of CO 2 formed = 0.616 g Mass of H 2 O formed = 0.108 g We know that CO 2 = C and H 2 O = 2H44 g 12 g 18 g 2g (iii) Percentage of O in the compound = 100 - (68.85 + 4.92)..