then the value of g(5) is
4
x = 3, it must be continuous at x = 3.
g(x) is differentiable at x = 3 \ m = k/4 Now, 3m + 2 = 2k yields m = 2/5, k = 8/5 g(5) = 5m + 2 = 4
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50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pKb of ammonia solution is 4.75, the pH of the mixture will be : Option: 1 3.75 Option: 2 4.75
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