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4 years ago

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Which element, represented by X, reacts with fluorine to produce the compound XF2?

2 answers:

Natalija [7]4 years ago

6 0

Answer:

This question is incomplete but the complete question is pasted below and the answer is 3

Question

Which element, represented by X, reacts with fluorine to produce the compound XF2?

(1) aluminum (3) magnesium

(2) argon (4) sodium

Explanation:

From the above question, X is unknown while F is fluorine.

For an unknown element to react with fluorine and the product will be XF2, the unknown element must have 2 electrons in its outermost shell and its ion must have a charge of 2+. Consider the equations below

X1+ + F1- ⇒ XF

X2+ + F1- ⇒XF2

X3+ + F1- ⇒XF3

From the options provided, aluminium ion has a charge of 3+, argon is a noble gas and hence has no charge, Magnesium has a charge of 2+ and sodium has a charge of 1+. Correspondingly, X is Magnesium and the compound formed will be MgF2.

NOTE:

Sodium will form NaF with flourine while aluminium will form AlF3 with fluorine. Argon will resist reaction with fluorine as it (argon) has achieved its octet configuration and hence stable.

Gala2k [10]4 years ago

3 0

The answer is magnesium.

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Assume the volume of the crew cabin is 74,000 L. The pressure is maintained at around 1.00 atm, ideally with an 80% nitrogen and

hjlf

Answer:

The number of moles of oxygen present in the crew cabin at any given time is 615.309 moles

Explanation:

The given parameters are;

Volume of the crew cabin = 74,000 L

Pressure of the crew cabin = 1.00 atm

Percentage of nitrogen in the mixture of gases in the cabin = 80%

Percentage of oxygen in the mixture of gases in the cabin = 20%

Temperature of the cabin = 20°C = 293.15 K

Therefore, volume of oxygen in the crew cabin = 20% of 74,000 L

Hence, volume of oxygen in the crew cabin = $\frac{20}{100} \times 74,000 \, L = 14,800 \, L$

From the universal gas equation, we have;

$n = \frac{P \times V}{R \times T}$

Where:

n = Number of moles of oxygen

P = Pressure = 1.00 atm

V = Volume of oxygen = 14,800 L

T = Temperature = 293.15 K

R = Universal Gas Constant = 0.08205 L·atm/(mol·K)

Plugging in the values, we have;

$n = \frac{1 \times 14,800 }{0.08205 \times 293.15 } = 615.309 \, moles$

The number of moles of oxygen present in the crew cabin at any given time = 615.309 moles.

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