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algol [13]
3 years ago
6

Iodine, I2, is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed. What is the temperature

in a 83.3-mL bulb that contains 0.392 g of I2 vapor at a pressure of 0.562 atm
Chemistry
1 answer:
Finger [1]3 years ago
3 0

Answer:

T = 3.75 K

Explanation:

As we know

PV = nRT

R =8.3144598 J. mol-1. K-1

P = 0.562 atm

V = 83.3 mL

moles in 0.392 g of I2 = 0.392/mass of I2 = 0.392 grams/253.8089 g/mol = 0.0015 moles

Substituting the given values, we get

0.562 atm * 83.3 *10^-3 L = 0.0015 moles * 8.3144598 J. mol-1. K-1 * T

T = 3.75 K

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What is the molarity of a solution that contains 87.75g of NaCI in 500.ml of solution
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M = 3.0 mol/L.

Explanation:

  • We can calculate the molarity of a solution using the relation:

<em>M = (mass x 1000) / (molar mass x V)</em>

  • M is the molarity "number of moles of solute per 1.0 L of the solution.
  • mass is the mass of the solute (g) (m = 87.75 g of NaCl).
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