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3 years ago

Because water molecules are polar and carbon dioxide molecules are nonpolar,

2 answers:

5 0

Because water molecules are polar and Carbon dioxide molecules are nonpolar : D. Water has higher boiling point than carbon dioxide does

Polar molecules have more bond compared to the non polar. which means it will take more energy to break polar molecules which resulted in higher boiling point

hope this helps

Gala2k [10]3 years ago

4 0

The correct is D.

Water is a polar molecule and it has polar bonds, which carry partially positive and partially negative charges. This polar bond increases the attraction between molecules of water and thus it requires a greater energy to break the bond between the molecules of water compare to carbon dioxide, which is a non polar molecule. Thus, water has a higher boiling point than carbon dioxide.

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How many grams of barium sulfate are produced if 25.34 mL of 0.113 M BaCl2 completely react given the reaction: BaCl2 (aq) + Na2

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<u>Answer:</u> The amount of barium sulfate produced in the given reaction is 0.667 grams.

<u>Explanation:</u>

To calculate the number of moles from molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of barium chloride = 0.113 M

Volume of barium chloride = 25.34 mL = 0.02534 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$0.113mol/L=\frac{\text{Moles of barium chloride}}{0.02534L}\\\\\text{Moles of barium chloride}=0.00286mol$

For the given chemical reaction:

$BaCl_2(aq.)+Na_2SO_4(aq.)\rightarrow BaSO_4(s)+2NaCl(aq.)$

By Stoichiometry of the reaction:

1 mole of barium chloride is producing 1 mole of barium sulfate.

So, 0.00286 moles of barium chloride will produce = $\frac{1}{1}\times 0.00286mol=0.00286mol$ of barium sulfate.

Now, to calculate the mass of barium sulfate, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of barium sulfate = 233.38 g/mol

Moles of barium sulfate = 0.00286 moles

Putting values in above equation, we get:

$0.00286mol=\frac{\text{Mass of barium sulfate}}{233.38g/mol}\\\\\text{Mass of barium sulfate}=0.667g$

Hence, the amount of barium sulfate produced in the given reaction is 0.667 grams

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Explanation:

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