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wlad13 [49]
3 years ago
15

Select the best single answer. Which of the following accurately lists compounds in order of increasing solubility in water? LiC

l, Br2, O2, methanol (CH3OH) methanol (CH3OH), LiCl, O2, Br2 O2, Br2, LiCl, methanol (CH3OH) LiCl, methanol (CH3OH), Br2, O2
Chemistry
1 answer:
Ann [662]3 years ago
4 0

Answer:

The Correct increasing order of solubility is O2 < Br2 < LiCl < Methanol (CH3OH)

Explanation:

Solubility of compounds or molecules are solely dependent on its inter molecular forces or bonding present in them.

Molecules with Hydrogen bonding usually very soluble in water. Ionic compounds are also very soluble in water because they form ions in solutions. Molecules that possess van der waal forces are usually insoluble in water because they are non-polar.

  • O2 (oxygen gas) and Br2 (bromine gas) have van der waal forces in them. Van der waal forces are stronger in Br2 (bromine gas) than O2 (oxygen gas) because Br2 has more number of electrons.
  • LiCl is ionic in nature which makes it dissolve in water readily. it easily forms its ions (Li+ and Cl- ) in solutions.
  • Methanol (CH3OH) has the highest solubility in water compared to LiCl, Br2 and O2 because it contains Hydrogen bonding which is strongest of all inter molecular forces.  

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d. To the left because Q > K_p

Explanation:

Hello,

In this case, for the given reaction:

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6 0
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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

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Molar mass of Na = 23 g/mol

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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

From the reaction, we conclude that

As, 2 mole of Na react to give 2 mole of NaCl

So, 0.217 mole of HCl react to give 0.217 mole of NaCl

Now we have to calculate the mass of NaCl

\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl

Molar mass of NaCl = 58.5 g/mole

\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g

Now we have to calculate the percent yield of this reaction.

Percent yield = \frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = \frac{5.24g}{12.7g}\times 100

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

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