Question

a student dissolves 8.00 g of $\ce{nh4no3}(s)$ in 100. g of $\ce{h2o}(l)$ in a coffee - cup calorimeter. based on the initial and final temperatures of the mixture in degrees celsius shown in the diagram above the laboratory setup, what is the calculated $\delta t$ of the water reported to the appropriate number of significant figures?\
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a $-5.500\\,^\circ\text{c}$\
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b $-5.50\\,^\circ\text{c}$\
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c $-5.5\\,^\circ\text{c}$\
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d $-6\\,^\circ\text{c}$

Explanation:

Step1: Recall the formula for temperature change

The formula for temperature change is $\Delta T = T_{final} - T_{initial}$. When dissolving $\ce{NH4NO3}$ in water, it's an endothermic process, so the final temperature will be lower than the initial, leading to a negative $\Delta T$.

Step2: Consider significant figures from given data

The mass of $\ce{NH4NO3}$ is $8.00\ \text{g}$ (3 significant figures), mass of water is $100.\ \text{g}$ (3 significant figures, the decimal after 100 indicates that the trailing zero is significant). For temperature change, we need to determine the appropriate number of significant figures. The initial and final temperatures (from the diagram, though not shown here, but typically in such problems, the temperature measurements would lead to a $\Delta T$ with two decimal places or two significant figures? Wait, no—wait, let's think about the options. The options have -5.50, -5.5, etc. Wait, when you calculate $\Delta T$, if the initial temperature is, say, 25.00 °C and final is 19.50 °C (for example), then $\Delta T = 19.50 - 25.00 = -5.50\ \text{°C}$. The given masses: 8.00 g (3 sig figs) and 100. g (3 sig figs). The temperature change's significant figures: the subtraction of temperatures—if initial is, say, 25.00 (4 sig figs) and final is 19.50 (4 sig figs), then $\Delta T$ is -5.50 (3 decimal? No, 25.00 - 19.50 = 5.50, so two decimal places? Wait, 25.00 has two decimal places, 19.50 has two decimal places, so the result of subtraction has two decimal places? Wait, no: 25.00 - 19.50 = 5.50, which is two decimal places? Wait, 25.00 is four significant figures, 19.50 is four. The difference is 5.50, which is three significant figures? Wait, 5.50 has three significant figures (the trailing zero after the decimal is significant). Wait, the mass of water is 100. g (three significant figures, because the decimal shows that the zero is significant), and $\ce{NH4NO3}$ is 8.00 g (three significant figures). So the temperature change should be reported with three significant figures? Wait, no—wait, the options: option B is -5.50 °C (three significant figures), option C is -5.5 (two), option A is -5.500 (four). Wait, but when you do the calculation, if the initial temperature is, for example, 25.0 °C (three sig figs) and final is 19.5 °C (three sig figs), then $\Delta T = 19.5 - 25.0 = -5.5$ (two sig figs? No, 25.0 has three, 19.5 has three, so the difference is 5.5 (two sig figs? Wait, 25.0 - 19.5 = 5.5, which is two significant figures? Wait, no: 25.0 is three, 19.5 is three, the difference is 5.5, which is two decimal places? No, 25.0 - 19.5 = 5.5, which is one decimal place? Wait, 25.0 has one decimal place, 19.5 has one decimal place, so the result has one decimal place. Wait, I think I messed up. Let's correct: the number of decimal places in subtraction is determined by the least number of decimal places in the numbers being subtracted. If initial temp is, say, 25.00 °C (two decimal places) and final is 19.50 °C (two decimal places), then $\Delta T = 19.50 - 25.00 = -5.50$ (two decimal places, three significant figures). The mass of $\ce{NH4NO3}$ is 8.00 g (three sig figs), water is 100. g (three sig figs). So the temperature change should have three significant figures? Wait, -5.50 has three significant figures (the 5, 5, and 0—wait, no: -5.50, the zero after the decimal is significant, so it's three significant figures. The option B is -5.50 °C, which has three significant figures, matching the significant figures from the masses (8.00 and 100. both have three sig figs). Option C is -5.5 (two sig figs), option A is -5.500 (four, which is too many), option D is -6 (one sig fig). So the correct answer should be B, -5.50 °C, because the data (8.00 g, 100. g) have three significant figures, and the temperature change calculation (subtraction) would result in three significant figures (or two decimal places, leading to -5.50).

Answer:

B. $-5.50\ ^{\circ}\text{C}$