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tamaranim1 [39]
3 years ago
5

For an experiment, you place 15 g of ice with a temperature of -10 C into a cup and label

Chemistry
2 answers:
melamori03 [73]3 years ago
6 0
The correct answer is c. Temperature is the average kinetic energy of a sample so if two samples have the same temperature they will also have the same average kinetic energy. I hope this helps. Let me know if anything is unclear.
Darina [25.2K]3 years ago
5 0

Average kinetic energy is given by the formula:

KE(average) = 3/2*k*T

where k = Boltzmann constant = 1.38*10⁻²³ J/K

T = temperature in Kelvin (K)

Thus, the average KE is dependent only on temperature and independent of mass.

Since the temp in Cup A and B remains the same, the average KE will also be the same.

Ans C)


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It can be done. Normally the boiling point of water is 100°C. It will boil at temperature greater than 100°C more quickly. Water can be boiled at 95°C but for that the atmospheric pressure of the water should be decreased which will decrease the boiling point of water.

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To boil water at 95°C, decrease the atmospheric pressure.

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A sample of 1.000 g of a compound containing carbon and hydrogen reacts with oxygen at elevated temperature to yield 0.692 g H₂O
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Answer :

(a) 1.000 g of compound containing carbon and hydrogen is, 0.922 g and 0.0769 g respectively.

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The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_y+O_2\rightarrow CO_2+H_2O

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Mass of H_2O=0.692g

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Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 3.381 g of carbon dioxide, \frac{12}{44}\times 3.381=0.922g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.692 g of water, \frac{2}{18}\times 0.692=0.0769g of hydrogen will be contained.

Thus, 1.000 g of compound containing carbon and hydrogen is, 0.922 g and 0.0769 g respectively.

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Mass carbon + Mass of hydrogen = 0.922 g + 0.0769 g = 0.999 g ≈ 1 g

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How many atoms of each element are in one molecule of this product.
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