Chemistry, 29.10.2019 19:31, yasameenakbari

From the calorimetric data, calculate δh for the reaction that occurs on mixing. assume that the calorimeter absorbs only a negligible quantity of heat, that the total volume of the solution is 100.0 ml, and that the specific heat and density of the solution after mixing are the same as that of pure water.

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answered: jennyn26

Explanation:

It is assumed that concentration of given $CuSO_{4}$ is 1.0 M and its volume is 50.0 ml or $50 \times 10^{-3}$ L (as 1 L = 1000 mL).

For KOH, concentration is 2.00 M and volume is 50.0 ml or $50 \times 10^{-3}$ L.

$T_{1}$ = $21.5^{o}C$ and $T_{2}$ = $27.7^{o}C$

Therefore, the given reaction equation will be as follows.

$CuSO_{4}(aq) + 2KOH (aq) \rightarrow Cu(OH)_{2}(s) + K_{2}SO_{4} (aq)$

Hence, calculate the moles of copper sulfate as follows.

Moles of $CuSO_{4} = molarity \times volume$

= $1.00 M \times 50.0 \times 10^{-3} L$

= 0.05 mol

Also, the number of moles of KOH will be as follows.

Moles of KOH = (molarity) × (volume)

= $2.00 M \times 50.0 \times 10^{-3} L$

= 0.10 mol

Therefore, moles $Cu(OH)_{2}$ formed = moles $CuSO_{4}$ consumed

0.5 × (moles of KOH consumed)

$0.5 \times 0.10 mol$

= 0.05 mol

It is given that volume of solution = 100.0 mL

Hence, mass of solution = (Volume of solution) × (density of solution)

= (100.0 mL) × (1.00 g/mL)

= 100.0 g

Also, heat absorbed by solution will be as follows.

q = $m \times C \times \Delta T$

= $100.0 g \times 4.184 J/g^{o}C \times (27.7 - 21.5)^{o}C$

= 2594.08 J

So, heat lost by reaction = -(Heat absorbed by solution)

As, heat lost by reaction = -(2594.08 J)

therefore, heat absorbed by reaction = 2594.08 J

Now, calculate the change in enthalpy as follows.

$\Delta H = \frac{\text{Heat lost by reaction}}{\text{moles Cu(OH)_{2} formed}}$

= $\frac{-2594.08 J}{0.05 mol)}$

= -51881.6 J/mol

or, = -51.9 kJ/mol (as 1 kJ = 1000 J)

Thus, we can conclude that $\Delta H$ for the given reaction that occurs on mixing is -51.9 kJ/mol.

answered: Guest

answer; i believe that the correct answer is (a) potassium(k)

answered: Guest

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is $1s^{2}2s^{2}2p^{6}3s^{1}$ and electronic configuration of phosphorous is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$ .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

$3na + p \rightarrow na_{3}p$

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.