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Sedaia [141]
4 years ago
9

A 270.0 gram piece of copper is dropped into 300.0 grams of water 24.00 °C. If the final temperature of water is 42.00 °C, what

was the initial temperature of the copper piece?
Specific heat of copper = 0.39 J/g °C

257 °C
272 °C
302 °C
313 °C
Chemistry
1 answer:
Papessa [141]4 years ago
4 0

Answer:

257

Explanation:

Just finished the test

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\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

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0.113mol/L=\frac{\text{Moles of barium chloride}}{0.02534L}\\\\\text{Moles of barium chloride}=0.00286mol

For the given chemical reaction:

BaCl_2(aq.)+Na_2SO_4(aq.)\rightarrow BaSO_4(s)+2NaCl(aq.)

By Stoichiometry of the reaction:

1 mole of barium chloride is producing 1 mole of barium sulfate.

So, 0.00286 moles of barium chloride will produce = \frac{1}{1}\times 0.00286mol=0.00286mol of barium sulfate.

Now, to calculate the mass of barium sulfate, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of barium sulfate = 233.38 g/mol

Moles of barium sulfate = 0.00286 moles

Putting values in above equation, we get:

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Hence, the amount of barium sulfate produced in the given reaction is 0.667 grams

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