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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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Answer : The initial concentration of HI and concentration of HI at equilibrium is, 0.27 M and 0.386 M  respectively.

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Let the initial concentration of HI be, C

The given equilibrium reaction is,

    H_2(g)+I_2(g)\rightleftharpoons 2HI(g)

Initially               0.11   0.11            C

At equilibrium  (0.11-x) (0.11-x)   (C+2x)

As we are given that:

Concentration of H_2 and I_2 at equilibrium = 0.052 M  = (0.11-x)

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By solving the terms, we get:

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Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate \rm CaCO_3?

Refer to a modern periodic table for relative atomic mass data:

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  • O: 15.999.

Formula mass of \rm CaCO_3:

M(\mathrm{CaCO_3})  = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}.

\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol.

How many moles of \rm CaCl_2 will be produced?

The coefficient in front of \rm CaCO_3 in the chemical equation is the same as that in front of \rm CaCl_2. That is:

\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1.

\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol.

What's the theoretical yield of calcium chloride? In other words, what's the mass of \rm 0.949184\;mol of \rm CaCl_2?

Again, refer to a periodic table for relative atomic data:

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M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}.

\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}.

What's the actual yield of calcium chloride?

\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%.

\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}.

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