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Rufina [12.5K]
3 years ago
6

Q)A certain mass of gas occupies a volume 2.5 L at 90atm. What pressure would the gas exert if it were placed in a 10 L containe

r at the same temperatur?​
Chemistry
1 answer:
rosijanka [135]3 years ago
4 0

Answer:

23 atm

Explanation:

Step 1: Given data

  • Initial volume (V₁): 2.5 L
  • Initial pressure (P₁): 90 atm
  • Final volume (V₂): 10 L
  • Final pressure (P₂): ?

Step 2: Calculate the final pressure of the gas

If we assume constant temperature and ideal behavior, we can calculate the final pressure of the gas using Boyle's law.

P₁ × V₁ = P₂ × V₂

P₂ = P₁ × V₁ / V₂

P₂ = 90 atm × 2.5 L / 10 L = 23 atm

As expected, since the volume increased, the pressure decreased.

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Assuming that the solubility of radon in water with 1 atm pressure of the gas over the water at 30 ∘C is 7.27×10−3M, what is the
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Answer:

K = 137.55 atm/M.

Explanation:

  • The relationship between gas pressure and the  concentration of dissolved gas is given by  Henry’s law:

<em>P = (K)(C)</em>

where P is the partial pressure of the gaseous  solute above the solution (P = 1.0 atm).

k is a constant (Henry’s constant).

C is the concentration of the dissolved gas (C = 7.27 x 10⁻³ M).

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3 years ago
Calculate the number of Na ions and Cl ions and total number of ions in 14.5g of NaCl.​
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Answer:

Number of Na ions in 14.5 g of NaCl is 1.49 × 10²³.

Number of Cl ions in 14.5 g of NaCl is 1.49 × 10²³.

Total number of ions =  1.49 × 10²³  +  1.49 × 10²³ = 2.98 × 10²³.

Explanation:

1 mole of any compound contains 6.023 × 10²³ molecules.

molecular weight of NaCl is 23 + 35.5 = 58.5 g.

so, 58.5 grams of NaCl makes 1 mole

⇒ 14.5 g of NaCl = \frac{14.5}{58.5} =  0.248 moles.

⇒ 0.248 mole contains 0.248 × 6.023×10²³ molecules

=  1.49 × 10²³ molecules.

And 1 molecule contains 1 Na ion and 1 Cl ion.

⇒ number of Na ions in 14.5 g of NaCl is 1.49 × 10²³.

⇒ number of Cl ions in 14.5 g of NaCl is 1.49 × 10²³.

Total number of ions =  1.49 × 10²³  +  1.49 × 10²³ = 2.98 × 10²³.

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