Carbon dioxide and helium which has a higher van der waals force of attraction between molecules

Chemistry

1 answer:

raketka [301]3 years ago

7 0

Answer:

Carbon dioxide

Explanation:

Neither helium nor carbon dioxide has a molecular dipole, so their strongest van der Waals attractive forces are London forces.

Helium is a small spherical atom with only a two electrons, so its atoms have quite weak attractions to each other.

CO₂ is a large linear molecule. It has more electrons than helium, so the attractive forces are greater. Furthermore, the molecules can align themselves compactly side-by-side and maximize the attractions (see below).

For example. CO₂ becomes a solid at -78 °C, but helium must be cooled to -272 °C to make it freeze (that's just 1 °C above absolute zero).

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3 years ago

A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi

Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of $Na$ = 5.0 g

Mass of $Cl_2$ = 10.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

First we have to calculate the moles of $Na$ and $Cl_2$ .

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$