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2 years ago

Which of these elements is a noble gas? A. hydrogen (H) B. barium (Ba) C. nitrogen (N) D. krypton (Kr) E. bromine (Br)

Chemistry

2 answers:

Citrus2011 [14]2 years ago

7 0

Answer: The correct answer is Option D.

Explanation:

Noble gases are defined as the gases which have stable electronic configuration and is highly nonreactive. These belong to Group 18 of the periodic table.

For the given options:

Option A: Hydrogen (H)

The electronic configuration of hydrogen = $1s^1$

This element belongs to Group 1 and is known as alkali metal.

Option B: Barium (Ba)

The electronic configuration of barium = $[Xe]6s^2$

This element belongs to Group 2 and is known as alkaline earth metal.

Option C: Nitrogen (N)

The electronic configuration of nitrogen = $1s^12s^22p^3$

This element belongs to Group 15 and is known as p-block element.

Option D: Krypton (Kr)

The electronic configuration of krypton = $[Ar]4s^23d^{10}4p^6$

This element belongs to Group 18 and is known as noble gas.

Option E: Bromine (Br)

The electronic configuration of bromine = $[Ar]4s^23d^{10}4p^5$

This element belongs to Group 17 and is known as halogen.

Hence, from the above information, the correct answer is Option D.

Novosadov [1.4K]2 years ago

5 0

Krypton is a noble gas! Hope this helps!

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As part of a soil analysis on a plot of land, a scientist wants to determine the ammonium content using gravimetric analysis wit

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Mass percentage of NH₄Cl = 3.54%

Mass percentage of K₂CO₃ = 1.01%

Explanation:

If a 200.0 mL aliquot produced 0.105 g of KB(C₆H₅)₄, then a 100.0 mL aliquot would produce 1/2 * 0.105 g = 0.0525 g of KB(C₆H₅)₄.

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Number of moles of NH₄B(C₆H₅)₄ in 0.2245 g = 0.2245 g/ 337.27 g/mol = 0.0006656 moles

In 500 ml solution, number of moles present = 0.0006656 * 500/100 = 0.003328 moles.

From equation of the reaction; mole ratio of NH₄⁺ and NH₄B(C₆H₅)₄ = 1:1

Similarly, mole ratio of NH₄⁺ and NH₄Cl = 1:1

Therefore, moles of NH₄Cl in 500 ml sample = 0.003328 moles

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From equation of the reaction; mole ratio of K⁺ and KB(C₆H₅)₄ = 1:1

Similarly, mole ratio of K⁺ and K₂CO₃ = 2:1

Therefore, moles of K₂CO₃ in 500 ml sample = 0.0007326/2 moles = 0.0003663 moles

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Calculate the volume occupied by 272g

cupoosta [38]

The answer for the following problem is mentioned below.

Therefore volume occupied by methane gas is 184.78 × 10^-3 liters

Explanation:

Given:

mass of methane( $CH_{4}$ ) = 272 grams

pressure (P) = 250 k Pa =250×10^3 Pa

temperature(t) = 54°C =54 + 273 = 327 K

Also given:

R = 8.31JK-1 mol-1 ,

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We know;

According to the ideal gas equation,

P × V = n × R × T

here,

n = m÷M

n =272 ÷ 16

n = 17 moles

Therefore,

250×10^3 × V = 17 × 8.31 × 327

V = ( 17 × 8.31 × 327 ) ÷ ( 250×10^3 )

V = 184.78 × 10^-3 liters

Therefore volume occupied by methane gas is 184.78 × 10^-3 liters

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