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baherus [9]
2 years ago
11

Explain the difference between heat and temperature. Does 1 L of water at 658°F have more, less, or the same quantity of energy

as 1 L of water at 65°C?
Chemistry
1 answer:
pogonyaev2 years ago
3 0

The answer is- The energy of 1 L water at temperature 347.78 °C have more energy as 1 L of water at temperature 65°C.

Heat is a type of energy that causes a person's body to feel hot or cold.

While the temperature of an object is a parameter that indicates how hot or cold the object is.

How is the temperature in degree Fahrenheit converted to degree celsius?

  • To convert the temperature in Fahrenheit to Celsius, subtract 32 and multiply by 5/9.

°C =( ^0F -32) * \frac{5}{9}

  • Now, heat is a form of energy that flows from hotter object to colder object and temperature indicates whether the object is hot or cold by measuring its average kinetic energy.
  • Now, the given temperature of 1 L water is 658 °F. This temperature in degree celsius is calculated as-

°C = (658 F-32) *\frac{5}{9} = 347.78 \° C

  • Now, higher the temperature, higher is the energy of water. Thus, the energy of 1 L water at 347.78 °C have more energy as 1 L of water at 65°C.

To learn more about heat and temperature, visit:

brainly.com/question/20038450

#SPJ4

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A 3.76 g sample of a compound consisting of carbon, hydrogen, oxygen, nitrogen, and sulfur was combusted in excess oxygen. This
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Answer: The empirical formula is C3H8S2NO.

Explanation: To determine the formula, find the porcentage of mass of each element in the sample:

1) For C (Molar Mass = 12u)

Note: molar mass of CO2 = 44g/mol

1 mol of CO2 = \frac{4,06}{44} = 0,092g/mol

In 1 mol of CO2, 1 mol of C, so mass of C is

mC = 0,092 . 12 = 1,104g

\frac{mC}{msample} . 100 = \frac{1,104}{3,76} . 100 = 29,4%

2) For H (Molar Mass = 1u)

Note: molar mass of H20 = 18g/mol

1 mol of H2O = \frac{2,22}{18} = 0,124 g/mol

In 1 mol of H2O, 2 mols of H, so mass is

mH = 0,124 . 2 . 1 = 0,248g

\frac{mH}{msample} . 100 = \frac{0,248}{3,76} . 100 = 6,6%

3) For S (Molar Mass = 32u)

Note: molar mass of SO3 = 80g/mol

1 mol of SO3 = \frac{3,73}{80} = 0,046g/mol

In 1 mol of SO3, 1 mol of S, so its mass is:

mS = 0,046 . 32 = 1,472g

\frac{mS}{msample} . 100 = \frac{1,472}{3,73} . 100 = 39,5%

4) For N (Molar Mass = 14u)

Note: molar mass of HNO3 = 63g/mol

1 mol of HNO3 = \frac{4,40}{63} = 0,07

In 1 mol of HNO3, 1 mol of N, so the mass is

mN = 0,07 . 14 = 0,98g

\frac{mN}{msample} . 100 = \frac{0,98}{8,53} . 100 = 11,5%

5) For O (Molar Mass = 16u).

O = 100% - (29,4%+6,6%+39,5%+11,5%)

O = 13%

It's determined the porcentage of the composition. Assuming that 100g of the compound, we calculte the mols of each:

Carbon = \frac{29,4}{12} = 2,45 mol

Hidrogen = 6,6 mol

Sulfur = 1,25 mol

Nitrogen = 0,82 mol

Oxygen = 0,81 mol

Divide each by 0,81, the empirical formula of the compound  is C3H8S2NO

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