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Maru [420]
3 years ago
10

Give the following for SO2 and BrF5:

Chemistry
1 answer:
frosja888 [35]3 years ago
6 0

Answer:

The given molecules are SO2 and BrF5.

Explanation:

Consider the molecule SO2:

The central atom is S.

The number of domains on S in this molecule is three.

Domain geometry is trigonal planar.

But there is a lone pair on the central atom.

So, according to VSEPR theory,

the molecular geometry becomes bent or V-shape.

Hybridization on the central atom is

sp^{2}.

Consider the molecule BrF5:

The central atom is Br.

The number of domains on the central atom is six.

Domain geometry is octahedral.

But the central atom has a lone pair of electrons.

So, the molecular geometry becomes square pyramidal.

The hybridization of the central atom is sp^{3} d^{2}.

The shapes of SO2 and BrF5 are shown below:

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Answer:

False

Explanation:

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Express the following quantity in scientific notation. The answer needs to have the correct number of significant figures. 21,30
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The combustion of 0.1240 kg of propane in the presence of excess oxygen produces 0.3110 kg of carbon dioxide. What is the limiti
nadezda [96]

Answer:

The limiting reactant is the propane gas, C₃H₈ while the percentage yield is 83.77%

Explanation:

Here we have

Propane gas with molecular formula C₃H₈, molar mass  = 44.1 g/mol combining with O₂ as follows

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Therefore, 1 mole of C₃H₈  combines with 5 moles of O₂ to produce 3 moles CO₂ and 4 moles of H₂O

Mass of propane = 0.1240 kg = 124.0 g

Number of moles of propane = mass of propane/(molar mass of propane)

The number of moles of propane = 124/44.1 = 2.812 moles

The molar mass of CO₂ = 44.01 g/mol

Mass of CO₂ = 0.3110 kg = 311.0 g

Therefore, number of moles of CO₂ = mass of CO₂/(molar mass of CO₂)

The number of moles of CO₂ = 311.0 kg/ 44.01 g/mol = 7.067 moles

Therefore, since 1 mole of propane produces 3 moles of CO₂, 2.812 moles of propane will produce 3 × 2.812 moles or 8.44 moles of CO₂

Therefore;

The limiting reactant is the propane gas, C₃H₈, since the oxygen is in excess

Hence

The \ percentage \ yield = \frac{Actual \, yield}{Theoretical \, yield} \times 100 = \frac{7.067}{8.44} \times 100 = 83.77 \%

The percentage yield = 83.77%.

7 0
3 years ago
How to go this problem
spin [16.1K]
Dccccccccccccccccccccccsssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss
4 0
3 years ago
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