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elena-14-01-66 [18.8K]
3 years ago
13

If 25.0 mL of 0.451 M NaOH solution is titrated with 0.253 M H2SO4, the flask at the endpoint will contain (besides the indicato

r phenolphthalein) as the principal components:
Chemistry
1 answer:
tigry1 [53]3 years ago
4 0

Answer:

H₂O, Na⁺, and SO₄²⁻

Explanation:

NaOH and H₂SO₄ are strong electrolytes. They will ionize completely .

Molecular equation:          2NaOH + H₂SO₄ ⟶ Na₂SO₄ + 2H₂O

Ionic equation: 2Na⁺ + 2OH⁻ + 2H⁺ + SO₄²⁻ ⟶  2Na⁺ + SO₄²⁻+ 2H₂O

  • The concentration of water is 55.5 mol·L⁻¹.
  • The concentration of Na⁺ is twice that of SO₄²⁻ .
  • Hydrolysis of SO₄²⁻ is negligible.

The principal components in the flask at the endpoint are H₂O, Na⁺, and SO₄²⁻.

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A 1.20-L container contains 1.10 g of an unknown gas at STP. What is the molecular weight of the unknown gas?
SOVA2 [1]

Answer:

M = 20.5 g/mol

Explanation:

Given data:

Volume of gas = 1.20 L

Mass of gas = 1.10 g

Temperature and pressure = standard

Solution:

First of all we will calculate the density.

Formula:

d = mass/ volume

d = 1.10 g/ 1.20 L

d = 0.92 g/L

Now we will calculate the molar  mass.

d = PM/RT

0.92 g/L = 1 atm × M / 0.0821 atm.L/mol.K ×273.15 K

M =  0.92 g/L × 0.0821 atm.L/mol.K ×273.15 K /  1 atm

M = 20.5 g/mol

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What did Becquerel expect to find when he developed the film from the photographic plates in the drawer? He expected to find one
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3 years ago
Nitrogen effuses through a pinhole 1.7 times as fast as another gaseous element under the same conditions. Estimate the other el
marysya [2.9K]

Answer:

80.92, Krypton

Explanation:

<u>What is effusion?</u>

• It is a process where gas escapes through a pinhole (a very small hole) into a region of low pressure or vacuum

<u>Graham's law of effusion of </u><u>gas</u>

• states that at a given constant temperature and pressure, the rate of effusion of gases is inversely proportional to the square root of their molar masses

\boxed{ \frac{Rate_1}{Rate_2} =  \sqrt{ \frac{M_2}{M_1} } }

<u>Calculations</u>

Nitrogen exist as N₂ at room temperature, thus its molar mass is 2(14)= 28.

Let the rate and molar mass of unknown gas be Rate₂ and M₂ respectively.

Since N₂ effuses 1.7 times as fast as the unknown gas,

Rate₁= 1.7(Rate₂)

\frac{Rate_1}{Rate_2} = 1.7

1. 7 =  \sqrt{ \frac{M_2}{28} }

Square both sides:

2.89  = \frac{M_2}{28}

Multiply both sides by 28:

2.89(28)= M₂

M₂= 80.92

<u>Identity of </u><u>gas</u>

The molar mass of 80.92 lies between Bromine and Krypton. However since Bromine exist as Br₂, the value of it's molar mass would be 159.8 instead. Hence, Bromine is eliminated.

If the gas is a diatomic element, the atomic weight is 80.92 ÷2= 40.46. Thus, we are now considering if Argon could be its identity. However, Argon is a noble gas and will not exist as a diatomic element. Argon is therefore eliminated too.

Thus based on the above reasoning, its probable identity is Krypton.

7 0
3 years ago
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