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Svetlanka [38]
3 years ago
9

A vessel contained N2, Ar, He, and Ne. The total pressure in the vessel was 1100 torr. The partial pressures of nitrogen, argon,

and helium were 110, 250, and 400 torr, respectively. The partial pressure of neon in the vessel was __________ torr.
A) 420 B) 340 C) 200 D) 280 E) 760
Chemistry
1 answer:
geniusboy [140]3 years ago
8 0

Answer:

Partial pressure Ne = 340 Torr

Option B

Explanation:

Gases contained in the vessel:

N₂, Ar, He, Ne

One of Dalton's law for gases determine this:

In a mixture of gases contained in a vessel, total pressure of the system must be the sum of partial pressure of each gas.

Total pressure = 1100 Torr

Let's replace:

Partial pressure N₂ + Partial pressure Ar + Partial pressure He + Partial pressure Ne  =  1100 Torr

Partial pressure Ne = 1100 Torr - Partial pressure N₂ - Partial pressure Ar  -Partial pressure He

Partial pressure Ne = 1100 Torr - 110 Torr - 250 Torr - 400 Torr

Partial pressure Ne = 340 Torr

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Wewaii [24]

Answer:

  • i) 5.00 mL of 1.00 M NaOH: before the equivalence point
  • ii) 50.0 mL of 1.00 M NaOH: before the equivalence point
  • iii) 100 mL of 1.00 M NaOH: at the equivalence point
  • iv) 150 mL of 1.00 M NaOH: after the equivalence point
  • v) 200 mL of 1.00 M NaOH: after the equivalence point

Explanation:

1. First calculate the number of mol acid in the 100 mL of 1.00 M HCl solution.

Equation:

  • Molarity = numbrer of moles of solute / volume of the solution in liters.

Thus,  you need to convert each volume from mL to liters, which is done dividing by 1,000.

Naming M the molarity, n the number of moles of solute (acid or base), and V the volume in liters:

M=n/v\implies n=M\times V=1.00M\times 0.100L=0.100mol

2. Now calculate the number of moles of NaOH for every condition (addition)

<u>i) 5.00 mL of 1.00 M NaOH</u>

n=0.00500liter\times 1.00M=0.00500molNaOH

Since the number of moles of NaOH added (0.00500mol) is less than the number of moles of acid in the solution (0.100mol), this is before equivalence point.

<u>ii) 50.0 mL of 1.00 M NaOH</u>

n=0.0500liter\times 1.00M=0.0500molNaOH

Since the number of moles of NaOH added (0.0500mol) is less than the number of moles of acid in the solution (0.100mol), this is before equivalence point.

<u />

<u>iii) 100 mL of 1.00 M NaOH</u>

n=0.100liter\times 1.00M=0.100molNaOH

Since the number of moles of NaOH added (0.100mol) is equal to the number of moles of acid in the solution (0.100mol), this is at the equivalence point.

<u>iv) 150 mL of 1.00 M NaOH</u>

n=0.150liter\times 1.00M=0.150molNaOH

Since the number of moles of NaOH added (0.150mol) is greater than the number of moles of acid in the solution (0.100mol), this is after the equivalence point.

<u>v) 200 mL of 1.00 M NaOH</u>

This is more volume of NaOH, then this is also after the equivalence point.

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