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3 years ago

Which of these statement(s) is/are true about a balloon filled with 1.00 mol N2 (g) at STP?

Chemistry

2 answers:

lara [203]3 years ago

7 0

Answer:

I. The balloon has a volume of 22.4L

III. The balloon contains 6.022x10^23 molecules.

Explanation:

At stp, it has been proven that 1mole of a gas occupy 22.4L.

Therefore, option (i) is correct.

The molar mass N2 = 14.01 x 2 = 28.02g/mol

Number of mole of N2 = 1 mole

Mass of N2 =..?

Mass = mole x molar Mass

Mass of N2 = 1 x 28.02 = 28.02g.

The mass content of the balloon is 28.02g, therefore, option (ii) is wrong.

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02x10^23 molecules. This implies that 1 mole of N2 also contains 6.02x10^23 molecules

Therefore, option (iii) is correct.

The correct options to the question are:

Option i and option iii

bogdanovich [222]3 years ago

4 0

Answer:

I and III are correct

Explanation:

Hello,

In this case, I. and III. are true since:

I. the STP conditions are referred to 273 K and 1 atm, therefore the volume is:

$V=\frac{nRT}{P}=\frac{1mol*0.082\frac{atm*L}{mol*K}*273K}{1atm}=22.4L$

II. In 1 mole of nitrogen (N₂) there are 28.02 g as it is diatomic, therefore this is false.

III. 1 mole of a substance, in this case nitrogen, equals 6.022x10²³ (Avogadro's number) particles, in this case molecules

Best regards.

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A type of plate boundary that is transform faults (easily seen where they cut at right angles to the mid-ocean ridges) A.) conve

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The answer is Option C (Divergent Plate Boundary)

Mapping efforts have shown that mid-ocean ridges are discontinuous structures that cut at right angles to its length at various transform faults. They typically demarcate the boundary between two tectonic plates, and are therefore called divergent plate boundaries.

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3 years ago

Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu

Monica [59]

Answer: The volume of the 1.224 M NaOH solution needed is 26.16 mL

Explanation:

In order to prepare the dilute NaOH solution, solvent is added to a given amount of the NaOH stock solution up to a final volume of 250.0 mL.

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M = n ÷ V

where n is the number of moles of the solute in the solution, and V is the volume of the solution.

Accordingly, the number of moles of the solute is given by

n = M x V

Now, let's designate the stock NaOH solution and the dilute solution as (1) and (2), respectively . The number of moles of NaOH in each of these solutions is:

n (1) = M (1) x V (1)

n (2) = M (2) x V (2)

As the amount of NaOH in the dilute solution is the same as in the volume taken from the stock solution,

n (1) = n (2)

and

M (1) x V (1) = M (2) x V (2)

For the stock solution, M (1) = 1.244 M, and V (1) is the volume needed. For the dilute solution, M (2) = 0,1281 M, and V (2) = 250.0 mL.

The volume of the stock solution needed, V (1), is calculated as follows:

V (1) = M (2) x V (2) ÷ M (1)

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3 years ago

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Answer:

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