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belka [17]
3 years ago
12

In which instance is a gas most likely to behave as an ideal gas?A.) At low temperatures, because the molecules are always far a

partB.) When the molecules are highly polar, because IMF are more likelyC.) At room temperature and pressure, because intermolecular interactions are minimized and the particles are relatively far apartD.) At high pressures, because the distance between molecules is likely to be small in relation to the size of the molecules
Chemistry
1 answer:
kiruha [24]3 years ago
3 0

Answer:

C.) At room temperature and pressure, because intermolecular interactions are minimized and the particles are relatively far apart.

Explanation:

For gas to behave as an ideal gas there are 2 basic assumptions:

  • The intermolecular forces (IMF) are neglectable.
  • The volume of the gas is neglectable in comparison with the volume of the container.

<em>In which instance is a gas most likely to behave as an ideal gas?</em>

<em>A.) At low temperatures, because the molecules are always far apart.</em> FALSE. At low temperatures, molecules are closer and IMF are more appreciable.

<em>B.) When the molecules are highly polar, because IMF are more likely.</em> FALSE. When IMF are stronger the gas does not behave as an ideal gas.

<em>C.) At room temperature and pressure, because intermolecular interactions are minimized and the particles are relatively far apart.</em> TRUE.

<em>D.) At high pressures, because the distance between molecules is likely to be small in relation to the size of the molecules.</em> FALSE. At high pressures, the distance between molecules is small and IMF are strong.

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Why does an exothermic reaction need activation energy?
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Weight of one mole of carbon = 12.01 g Weight of one mole of oxygen = 16.00 g The molecular weight (gram formula weight) for CO
murzikaleks [220]

Answer:

28.01g

Explanation:

Given the weight of one mole of Cabon as 12.01g and that of oxygen as 16.00g.

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4 0
4 years ago
What would be the freezing point of a solution that has a molality of 1.324 m which was prepared by dissolving biphenyl (C12H10)
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The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

A solution is prepared by dissolving biphenyl into naphthalene. We can calculate the freezing point depression (ΔT) for naphthalene using the following expression.

\Delta T = i \times Kf \times m =   1 \times 6.90 \°C/m  \times 1.324m = 9.14  \°C

where,

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The normal freezing point of naphthalene is 80.26 °C. The freezing point of the solution is:

T = 80.26 \° C - 9.14 \° C = 71.12 \° C

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

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3 years ago
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3 years ago
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