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Kipish [7]
3 years ago
12

A 0.532 mol sample of SO2 gas requires 52.3 s to effuse through a tiny hole. Under the same conditions, how long will it take 0.

532 mol of Ar gas to effuse?
Chemistry
1 answer:
scoundrel [369]3 years ago
3 0

Answer:

41.3 s

Explanation:

Let t₁ represent the time taken for SO₂ to effuse.

Let t₂ represent the time taken for Ar to effuse.

Let M₁ represent the molar mass of SO₂

Let M₂ represent the molar mass of Ar

From the question given above,

Time taken (t₁) for SO₂ = 52.3 s

Time taken (t₂) for Ar =?

Molar mass (M₁) of SO₂ = 32 + (16×2) = 32 + 32 = 64 g/mol

Molar mass (M₂) of Ar = 40 g/mol

Finally, we shall determine the time taken for Ar to effuse by using the Graham's law equation as shown below:

t₂ / t₁ = √(M₂ / M₁)

t₂ / 52.3 = √(40 / 64)

t₂ / 52.3 = √0.625

t₂ / 52.3 = 0.79

Cross multiply

t₂ = 52.3 × 0.79

t₂ = 41.3 s

Thus, the time taken for the amount of Ar to effuse is 41.3 s

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I believe it is useful as it shows how electrons are distributed in the shells (energy levels) of an iron atom. An atom is the smallest particle of an element that can take part in a chemical reaction. It contains the nucleus and the shells (energy levels). the nucleus contains protons and neutrons while the shells contain electrons.
7 0
3 years ago
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What mass of NaCl is dissolved in 150 g of water in a .050<br> msolution?
mart [117]

Answer:

0.4383 g

Explanation:

Molality is defined as the moles of the solute present in 1 kg of the solvent.

It is represented by 'm'.

Thus,  

Molality\ (m)=\frac {Moles\ of\ the\ solute}{Mass\ of\ the\ solvent\ (kg)}

Given that:

Mass of solvent, water = 150 g = 0.15 kg ( 1 g = 0.001 g )

Molality = 0.050 m

So,

0.050=\frac {Moles\ of\ the\ solute}{0.15}

Moles = 0.050\times 0.15\ mol= 0.0075\ mol

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6 0
4 years ago
1Calculate the density of an object that has a mass of 84.7g and a volume of 59.3 cm3
Iteru [2.4K]

Answer:

<h2>1.43 g/cm³</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 84.7 g

volume = 59.3 cm³

We have

density =  \frac{84.7}{59.3}  \\  = 1.428330...

We have the final answer as

<h3>1.43 g/cm³</h3>

Hope this helps you

7 0
3 years ago
A 5% Dextrose solution is diluted with sterile water. The volume of the sterile water added is 40% of the volume of the Dextrose
baherus [9]

Answer:

4.90% is approximately the new Dextrose concentration.

Explanation:

Volume by Volume percent is given by ;

(v/v)\%=\frac{\text{Volume of solute}}{\text{Volume of solution}}\times 100

Volume percentage of dextrose solution = 5%

In 100 mL of solution 5 ml of dextrose is present.

Now, volume sterile water added was equal to the 40% of volume of dextrose volume.

So, volume of the sterile water added = \frac{40}{100}\times 5 ml = 2 mL

Total volume of the solution after addition of water = 100 mL + 1 mL = 102 mL

New concentration of dextrose will be;

(v/v)\%=\frac{\text{Volume of solute}}{\text{Volume of solution}}\times 100

\frac{5 mL}{102 mL}\times 100=4.90\%

4.90% is approximately the new Dextrose concentration.

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3 years ago
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AlekseyPX

Answer:

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D and 1

E and 2

A and 3

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B and 6

Explanation:

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