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3 years ago

How can you determine the specific heat of a metal using a calorimeter

Chemistry

1 answer:

zhannawk [14.2K]3 years ago

4 0

Answer:

One can determine the specific heat of the metal through using the clarimeter, water, thermometer and using heat equations.

Explanation:

You can learn about heat effects and calorimetery through a simple experiment by boiling water and heating up the metal in it. Then, pour it into your calorimeter and the heat will flow from the metal to the water. The two equlibria will meet: the metal will loose heat into its surroundings (the water) and teh water will absorb the heat. The heat flow for the water is the same as it is for the metal, the only difference being is the negative sign indicating the loss of the heat of the metal.

In terms of theromdynamics, we can deteremine the heat flow for the metal becasue it would be equal to the mangnitued but opposite in direction. Thus, we can say that the specific heat of water qH2O = -qmetal.

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Hydrogen peroxide can be prepared by the reaction of barium peroxide with sulfuric acid according to the reaction BaO2(s)+H2SO4(

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Answer:

VH2SO4 = 145.3 mL

Explanation:

Mw BaO2 = 169.33 g/mol

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Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH ( aq ) . Calculate t

Sedbober [7]

Answer:

Approximately $6.30\times 10^{-3}\;\rm mol$ .

Explanation:

The gallium here is likely to be produced from a $\rm NaGaO_2\, (aq)$ solution using electrolysis. However, the problem did not provide a chemical equation for that process. How many electrons will it take to produce one mole of gallium?

Note the Roman Numeral " $\mathtt{(III)}$ " next to $\rm Ga$ . This numeral indicates that the oxidation state of the gallium in this solution is equal to $+3$ . In other words, each gallium atom is three electrons short from being neutral. It would take three electrons to reduce one of these atoms to its neutral, metallic state in the form of $\rm Ga\, (s)$ .

As a result, it would take three moles of electrons to deposit one mole of gallium atoms from this gallium $\mathtt{(III)}$ solution.

How many electrons are supplied? Start by finding the charge on all the electrons in the unit coulomb. Make sure all values are in their standard units.

$t = \rm 80.0\; min = 80.0\; min \times 60\;s \cdot min^{-1} = 4800\; s$ .

$Q = I \cdot t = \rm 0.380 \; A \times 4800 \; s = 1.824\times 10^3\; C$ .

Calculate the number of electrons in moles using the Faraday's constant. This constant gives the size of the charge (in coulombs) on each mole of electrons.

$\begin{aligned} n(\text{electrons}) &= \frac{Q}{F} \cr &= \rm \dfrac{1.824\times 10^3\; C}{96485.332\; C \cdot mol^{-1}}\cr &\approx \rm 1.89\times 10^{-2}\; mol \end{aligned}$ .

It takes three moles of electrons to deposit one mole of gallium atoms $\rm Ga\, (s)$ . As a result, $\rm 1.89\times 10^{-2}\; mol$ of electrons would deposit $\displaystyle \rm \frac{1}{3}\times 1.89\times 10^{-2}\; mol \approx 6.30\times 10^{-3}\; mol$ of gallium atoms $\rm Ga\, (s)$ .

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3 years ago