Question

2naoh + h₂so₄ → 2h₂o + na₂so₄50.0 ml of 0.50 m h₂so₄ is added to 60.0 ml of 0.80 m naoh. the resulting temperature change is an increase of 1.20 °c.\\(c_{sol} = 4.20 j/g °c\\) \\(d_{sol} = 1.05 g/ml\\) \\(c_{cal} = 4.50 j/°c\\)\\(mass_{sol} = 116 g\\)what is the heat of reaction?remember: \\(q_{rxn} = -(q_{sol} + q_{cal})\\)\\(q_{rxn} = ? j\\)enter either a + or - sign and the magnitude.

Explanation:

Step1: Calculate \( q_{\text{sol}} \)

Use the formula \( q = mc\Delta T \). Here, \( m = 116\ \text{g} \), \( c = 4.20\ \text{J/g}\ ^{\circ}\text{C} \), \( \Delta T = 1.20\ ^{\circ}\text{C} \).
\( q_{\text{sol}} = 116\ \text{g} \times 4.20\ \text{J/g}\ ^{\circ}\text{C} \times 1.20\ ^{\circ}\text{C} \)
\( q_{\text{sol}} = 116 \times 4.20 \times 1.20 \)
\( q_{\text{sol}} = 574.56\ \text{J} \)

Step2: Calculate \( q_{\text{cal}} \)

Use the formula \( q = c\Delta T \). Here, \( c = 4.50\ \text{J/}^{\circ}\text{C} \), \( \Delta T = 1.20\ ^{\circ}\text{C} \).
\( q_{\text{cal}} = 4.50\ \text{J/}^{\circ}\text{C} \times 1.20\ ^{\circ}\text{C} \)
\( q_{\text{cal}} = 5.40\ \text{J} \)

Step3: Calculate \( q_{\text{rxn}} \)

Use the formula \( q_{\text{rxn}} = -(q_{\text{sol}} + q_{\text{cal}}) \).
First, find \( q_{\text{sol}} + q_{\text{cal}} = 574.56\ \text{J} + 5.40\ \text{J} = 579.96\ \text{J} \)
Then, \( q_{\text{rxn}} = -579.96\ \text{J} \) (approximately -580 J when rounded)

Answer:

\(-580\) (or more precisely \(-579.96\))