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Andreas93 [3]
2 years ago
8

The VSEPR model was developed before any xenon compounds had been prepared. Thus, these compounds provided an excellent test of

the model’s predictive power. What would you have predicted for the shapes of XeF₂, XeF₄, and XeF₆?
Chemistry
1 answer:
Natalka [10]2 years ago
6 0

The maximum amount of XeF4 that could be produced is 0.5 moles.

XeF4 = Xe (g) 2 F2 (g) (g)

Xe and F2 have a mole ratio of 1:2. Because of this, the reaction would be limited by F2 when there is 1 mole of Xe and 1 mole of F2.

<h3>What is mole ratio?</h3>

The mole ratio is the ratio of any two compounds' mole amounts that are present in a balanced chemical reaction.

A comparison of the ratios of the molecules required to accomplish the reaction is given by the balancing chemical equation.

A mole ratio is a conversion factor used in chemical reactions to link the mole quantities of any two compounds. A conversion factor's numbers are derived from the balanced chemical equation's coefficients.

To learn more about mole ratio from the given link:

brainly.com/question/14425689

#SPJ4

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You have a solution of 600 mg of caffeine dissolved in 100 mL of water. The partition coefficient for aqueous caffeine extracted
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Answer:

159 mg caffeine is being extracted in 60 mL dichloromethane

Explanation:

Given that:

mass of caffeine in 100 mL of water =  600 mg

Volume of the water = 100 mL

Partition co-efficient (K) = 4.6

mass of caffeine extracted = ??? (unknown)

The portion of the DCM = 60 mL

Partial co-efficient (K) = \frac{C_1}{C_2}

where; C_1= solubility of compound in the organic solvent and C_2 = solubility in aqueous water.

So; we can represent our data as:

K=(\frac{A_{(g)}}{60mL} ) ÷ (\frac{B_{(mg)}}{100mL} )

Since one part of the portion is A and the other part is B

A+B = 60 mL

A+B = 0.60

A= 0.60 - B

4.6=(\frac{0.6-B(mg)}{60mL} ) ÷ (\frac{B_{(mg)}}{100mL})

4.6 = \frac{(\frac{0.6-B(mg)}{60mL} )}{(\frac{B_{(mg)}}{100mL})}

4.6 × (\frac{B_{(mg)}}{100mL}) = (\frac{0.6-B(mg)}{60mL} )

4.6 B *\frac{60}{100} = 0.6 - B

2.76 B = 0.6 - B

2.76 + B = 0.6

3.76 B = 0.6

B = \frac{0.6}{3.76}

B = 0.159 g

B = 159 mg

∴ 159 mg caffeine is being extracted from the 100 mL of water containing 600 mg of caffeine with one portion of in 60 mL dichloromethane.

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Reduction on the copper cathode: Cu²⁺(aq) + 2e⁻ → Cu(s).

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