Question

I can't figure out how to do question 5 (part 1 and 2) and question 6, could you provide explanation
Question #4 One way to limit the choices for your unknown is to consider the sign of AHgoln. Considering your results from Table 8-3 and the table of heats of solution provided below, which of the following are possibilities for your salt considering only the sign? mass of solid mass of water. Tinitial Trnal Salt Trial #1 5.029 9 5 0.0 g 21.5 °C 27.3 °c NazC03 Trial # 2 5.089 g 50.0 9 21.5°C 27.7 °C KCI NH4Cl AH soln (1/9) -267 231 305 true: Na coz false: KC false: NH40 Question #5 Use your data from Table 8-3 to determine AHsoln for your unknown salt. Part 1: Enter the results of Trial #1 and Trial #2 for AHgoln in the table below. If you calculate the value of AHsoln as less than zero, make sure to include the negative sign. AHsoln (Trial #1) AH soln (Trial #2) J/9 Sutinit Part 2: Enter the average AHsoin for your two trials in the table below. If you calculate the average of AHsoln as less than zero, make sure to include the negative sign. AHgoln (average) 19

Answer 1

Q5. part 1

For trial 1

The experimental value for $\Delta$ _solH can be calculated by the following formula;

Q = m.c. $\Delta$ T; where Q is the heat released or absorbed, m is the mass of water, c is specific heat capacity of water, and  $\Delta$T is the change in temperature (T_i - T_f),

$\therefore$ Q = 50.0 g $\times$ 4179.6 J/kg. K $\times$ (21.5 - 27.3) K $\times$ (1kg/1000g) = -1212.084 J

[Note: (i) c for water = 4179.6 J/kg.K; (ii)The kg in 'c' is changed to grams by multiplying with the conversion factor]

Now, $\Delta$ _solH in J/g for trial 1 = Q/(mass of solid) = (-1212.084)/(5.029g) = -241.02 J/g

For trial 2

Following exactly the same procedure for trial 2

$\Delta$_solH in J/g for trial 2 = -254.60 J/g

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Q5. part 2

$\Delta$_solH (average) = [$\Delta$_solH(trial 1) + $\Delta$ _solH (trial 2)]/2 = -247.81 J/g

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Q6.

The salt is clearly Na2CO3 whose given value of $\Delta$ _solH = -267 J/g (Theoretical value)

% error = [(Theoretical value - Experimental value or average)/Theoretical value] $\times$ 100 = 7.19 %