When 400 mL of 0.2 M H_2SO_4 solution is m | vedclass

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(B) Step $1$: Calculate moles of $H^+$ and $OH^-$.$n_{H^+} = 2 	imes 	ext{Molarity} 	imes 	ext{Volume (L)} = 2 	imes 0.2 	imes 0.4 = 0.16 \ mol$

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$82$

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(B) Step $1$: Calculate moles of $H^+$ and $OH^-$.$n_{H^+} = 2 	imes 	ext{Molarity} 	imes 	ext{Volume (L)} = 2 	imes 0.2 	imes 0.4 = 0.16 \ mol$.$n_{OH^-} = 1 	imes 	ext{Molarity} 	imes 	ext{Volume (L)} = 0.1 	imes 0.6 = 0.06 \ mol$.Step $2$: Identify the limiting reagent. Since $n_{OH^-} Step $3$: Calculate heat released $(q)$.$q = n_{OH^-} 	imes |\Delta_{r}H| = 0.06 \ mol 	imes 57.1 	imes 10^3 \ J \ mol^{-1} = 3426 \ J$.Step $4$: Calculate temperature change $(\Delta T)$.Total mass of solution = $(400 + 600) \ mL 	imes 1.0 \ g \ mL^{-1} = 1000 \ g$.$q = m 	imes c 	imes \Delta T \Rightarrow 3426 = 1000 	imes 4.18 	imes \Delta T$.$\Delta T = \frac{3426}{4180} \approx 0.8196 \ K$.Expressing in $10^{-2} \ K$: $0.8196 	imes 100 	imes 10^{-2} \ K = 81.96 	imes 10^{-2} \ K$.Rounding to the nearest integer,we get $82 	imes 10^{-2} \ K$.

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