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3 years ago

Describe how you would use a calorimeter to determine the specific heat capacity of a metal

Chemistry

2 answers:

Ymorist [56]3 years ago

4 0

Answer: (see explanation part)

Explanation:

A piece of solid material of specific heat c unknown is weighed with a balance, m resulting in its mass. The piece is put in almost boiling water at temperature T.

M grams of water are put on the calorimeter, it is stirred and after a short time, its temperature T0 is measured.

The solid part is quickly deposited in the calorimeter. It is stirred, and after a certain time the equilibrium temperature Te is reached.

The data is recorded and c is cleared of the formula that's attached (please take a look).

vovikov84 [41]3 years ago

3 0

<span>My only guess is obtain a metal and heat it in a boiling water bath (of known temperature) this will be your initial temperature. Now obtain a calorimeter cup with water of known temperature as well. Place the metal into the calorimeter cup and record the temperature after 5 minutes. You now have delta T, mass of the metal, and Q. Solve for C.

Hope this helps xox :)</span>

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melamori03 [73]

Answer:

Option D. pH= 1.3 strong acid

Explanation:

From the question given:

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3 years ago

One of the reactions that occurs in a blast furnace, in which iron ore is converted to cast iron, is Fe2O3 + 3CO → 2Fe + 3CO2 Su

Tpy6a [65]

Answer : The percent purity of $Fe_2O_3$ in the original sample is 87.94 %

Explanation :

The given balanced chemical reaction is:

$Fe_2O3+3CO\rightarrow 2Fe+3CO_2$

First we have to calculate the mass of Fe.

$\text{Moles of }Fe=\frac{\text{Mass of }Fe}{\text{Molar mass of }Fe}$

Molar mass of Fe = 55.8 g/mole

$\text{Moles of }Fe=\frac{1.79\times 10^3kg}{55.8g/mole}=\frac{1.79\times 10^3\times 1000g}{55.8g/mole}=3.15\times 10^4mole$

Now we have to calculate the moles of $Fe_2O_3$

From the balanced chemical reaction we conclude that,

As, 2 moles of Fe produced from 1 mole of $Fe_2O_3$

So, $3.15\times 10^4mole$ of Fe produced from $\frac{3.15\times 10^4}{2}=15750$ mole of $Fe_2O_3$

Now we have to calculate the mass of $Fe_2O_3$

$\text{ Mass of }Fe_2O_3=\text{ Moles of }Fe_2O_3\times \text{ Molar mass of }Fe_2O_3$

Molar mass of $Fe_2O_3$ = 159.69 g/mole

$\text{ Mass of }Fe_2O_3=(15750moles)\times (159.69g/mole)=2.515\times 10^6g=2.515\times 10^3kg$

Now we have to calculate the percent purity of $Fe_2O_3$ in the original sample.

Mass of original sample = $2.86\times 10^3kg$

$\text{Percent purity}=\frac{\text{Mass of }Fe_2O_3}{\text{Mass of sample}}\times 100$

$\text{Percent purity}=\frac{2.515\times 10^3kg}{2.86\times 10^3kg}\times 100=87.94\%$

Therefore, the percent purity of $Fe_2O_3$ in the original sample is 87.94 %

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3 years ago

The change in the standard Gibbs free energy (ΔGº) for the dissociation of nitrous acid (HNO2) at 298 K is 19.09 kJ. If the pH o

Fynjy0 [20]

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zimovet [89]

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