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sesenic [268]
3 years ago
9

A 5.00 liter gas sample is collected at a temperature and pressure of 27.0 degree C and 1.20 atm. It is desired to transfer the

gas to a 3.00 liter container at a pressure of 1.00 atm. What must be the temperature of the gas in the 3.00 liter container?
a) 417 K
b) 216 K
c) 13.5 degree C
d) 150 K
Chemistry
1 answer:
kipiarov [429]3 years ago
3 0

Answer:

The temperature of the gas in the 3.00 liter container, must be 150K

Explanation:

Let's apply the Ideal Gas Law, to find out the moles

P . V = n . R . T

1,20 atm . 5L = n . 0,082 L.atm/mol.K . 300 K

(1,20 atm . 5L) /  (0,082 mol.K/L.atm . 300 K) = n

6/24,6 mol = n = 0,244 moles

We have the moles now, so let's find the temperature in our new conditions.

P . V = n . R . T

1 atm . 3L = 0,244 moles . 0,082 L.atm/mol.K . T° in K

(1 atm . 3L / 0,244 moles . 0,082 mol.K/L.atm) = T° in K

3/20,008 K = T° in K = 150K

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When an ionic compound such as sodium chloride (NaCl) is placed in water, the component atoms of the NaCl crystal dissociate int
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A) NaCl          ionic            ions Na⁺ and Cl⁻        1.0            1.0 × 1 × 2 = 2

B) NaCl          ionic            ions Na⁺ anc Cl⁻        0.5           0.5 × 1 × 2 = 1

C) Glucose    covalent     molecules                   0.5          0.5 × 1 × 1 = 0.5

D) Glucose    covalent     molecules                   1.0           1.0 × 1  × 1 = 1

Therefore, the rank in increasing number of particles is for the list of solutions given is: C < B = D < A, which means that the solution expected to contain the greatest number of solute particles is the solution A) 1 L of 1.0 M NaCl.

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