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Doss [256]
3 years ago
9

Which choice is a mixture?

Chemistry
1 answer:
Irina18 [472]3 years ago
6 0

Answer:

air (N2 mixed with O2 and CO2)

Explanation:

It's a mixture of different gases.

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8 valence electrons :)
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You collect a sample of gases from an indoor pool area. The sample contains air and water vapor. The total pressure is 100.18 ki
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C. 96.82 kPa because to find the amount of pressure the air is causing you need to subtract the amount of pressure the water vapor is causing because the only two gasses making up the air in the pool area are air and water vapor.

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2. (10 Points) If the experiment had continued with you adding an 8 kg weight to
vaieri [72.5K]

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number 6: because the temperature and the amount of gas don’t change, these terms don’t appear in the equation. What Boyle’s law means is that the volume of a mass of gas is inversely proportional to its pressure. This linear relationship between pressure and volume means doubling the volume of a given mass of gas decreases its pressure by half.

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The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera
torisob [31]

Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = \frac{0.570 g}{122.12 g/mol}=0.004667 mol

Energy released by 0.004667 moles of benzoic acid on combustion:

Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

Q=C\times \Delta T

15,066.8 J=C\times 2.053^oC

C=7,338.92 J/^oC

Glucose:

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

Moles of glucose = \frac{2.900 g}{180.16 g/mol}=0.016097 mol

Energy released by the 0.016097 moles of calorimeter  combustion:

Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

Q'=C\times \Delta T'

44,749.1 J=7,338.92 J/^oC\times \Delta T'

\Delta T'=6.097^oC

The temperature change from the combustion of the glucose is 6.097°C.

6 0
3 years ago
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