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agasfer [191]
3 years ago
14

A balloon vendor at a street fair is using a tank of helium to fill her balloons. The tank has a volume of 113.0 L and a pressur

e of 122.0 atm at 25.0 °C. After a while, she notices that the valve has not been closed properly. The pressure had dropped to 112.0 atm (The tank is still at 25.0 °C).
Chemistry
1 answer:
lisov135 [29]3 years ago
7 0
Given Data:
                   P₁  =  122 atm

                   P₂  =  112 atm

                   V₁  =  113 L

                   V₂  = ?

Solution:
              Let suppose the gas is acting ideally. According to Ideal gas equation, keeping the temperature constant,

                                  P₁ V₁  =  P₂ V₂
Solving for V₂,
                                      V₂  =  P₁ V₁ / P₂
Putting Values,
                                      V₂  =  (122 atm × 113 L) ÷ 112 atm
                 
                                      V₂  =  113 L

Converting Volume to Moles,
 As,
                         22.4 L is occupied by  =  1 mole of He gas
So,
                         113 L will occupy        =  X moles of He gas

Solving for X,
                                     X  =  (113 L × 1 mole) ÷ 22.4 L

                                     X  =  5.04 Moles of He 
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Explanation:

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It is logical to say that, because these sort of water have ions. Remember that colligative properties depends on the solute particles.

T° boiling solution - T° pure solvent = Kb . m . i

In the freezing point depression, we have the oppossite of boiling point elevation. Freezing point of solution is lower than pure solvent, according to this:

T° freezing pure solvent - T° freezing solution = Kc . m . i

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A) CH3CH2SH

Explanation:

Dispersion forces are weak attractions found between non-polar and polar molecules. The attractions here can be attributed to the fact that a  non-polar molecule sometimes become polar because the constant motion of its electrons may lead to an uneven charge distribution at an instant. If this happens, the molecule has a temporary dipole. This dipole can induce the neighbouring molecules to be distorted and form dipoles as well. The attractions between these dipoles constitute the Dispersion Forces.

Therefore; the greater the molar mass of a compound or molecule, the higher the Dispersion Force. This implies that the compound or molecule with the highest molar mass have the largest dispersion forces.

Now; for option (A)

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= (12 + (1 × 3 ) +12 + (1 ×2) + 32+1)

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= 12 + (1 × 4)

= 12 + 4

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= 12 + ( 1 × 3 ) + 12 + ( 1 × 3)

= 12 + 3 + 12 + 3

= 30 g/mol

Thus ; option (A) has the highest molar mass, as such the largest dispersion force is A) CH3CH2SH

7 0
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