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Alinara [238K]
4 years ago
12

A laboratory technician drops a 0.0850 kg sample of unknown material, at a temperature of 100.0∘C, into a calorimeter. The calor

imeter can, initially at 19.0∘C, is made of 0.150 kg of copper and contains 0.200 kg of water. The final temperature of the calorimeter can is 26.1∘C.
Chemistry
1 answer:
GREYUIT [131]4 years ago
5 0

Answer:

1013.32 J/kg.K

Explanation:

The heat transferred by a changing in temperature without phase change can be calculated by:

Q = m*c*ΔT

Where m is the mass, c is the specific heat, and ΔT is the change in temperature (final - initial).

The values of c for water and copper can e found in thermodynamics tables:

cwater = 4.19x10³ J/kg.K

ccopper = 0.39x10³ J/kg.k

By the conservation of energy:

Qwater + Qcopper + Qmaterial = 0

0.200*4.19x10³*(26.1 - 19.0) + 0.150*0.39x10³*(26.1 - 19.0) + 0.085*c*(26.1 - 100) = 0

5949.8 + 415.35 - 6.2815c = 0

6.2815c = 6365.15

c = 1013.32 J/kg.K

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Explanation:

We have to find the number of molecules that are present in 0.54 g of Ca(NO₃)₂.

First we have to convert the mass of our sample into moles of Ca(NO₃)₂. We will use the molar mass of Ca(NO₃)₂ to do that.

molar mass of Ca = 40.08 g/mol

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molar mass of Ca(NO₃)₂ = 164.10 g/mol

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moles of Ca(NO₃)₂ = 0.54 g * 1 mol/(164.10 g)

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According to Avogadro's number there are 6.022 *10^23 molecules in 1 mol of molecules. We can use that relationship to find the number of molecules that are present in our sample.

6.022 *10^23 molecules = 1 mol

molecules of Ca(NO₃)₂ = 0.00329 moles * 6.022 *10^23 molecules/(1 mol)

molecules of Ca(NO₃)₂ = 2.0 * 10^21 molecules

Answer: there are 2.0 * 10^21 molecules of Ca(NO₃)₂ in 0.54 g of it.

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