Suppose you mix 200.0 mL of 0.200 M RbOH(aq) with 100.0 mL of 0.400 M HBr...

Question

Question

Suppose you mix 200.0 mL of 0.200 M RbOH(aq) with 100.0 mL of 0.400 M HBr (aq) in a coffee cup calorimeter. The initial tempe

Answer 1

Answer #1

Answer

$\Delta H$ = - 2.2 kJ

Explanation

q = m*c* $\Delta T$

m = 300 g [ $\because$ density of solution is 1.0 g/mL, for 300 mL, m = 300 g, Total volume =200mL+100mL=300mL]

c = 4.18 J/g.C

$\Delta T$ = 1.78 C [ $\because$ $\Delta T$ = 26.18 - 24.4 = 1.78 C]

q = 300*4.18*1.78

q = 2232.12 J

$\Delta H$ = 2.232 kJ

Since the temperature is increased, the reaction is exothermic. Enthalpy of an exothermic raction is negative.

$\Delta H$ = - 2.232 kJ

$\Delta H$ = - 2.2 kJ [in one decimal point]

Answer 2

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