Chemistry, 18.11.2019 01:31 brittainpolk453

A4.05 sample of an unknown gas at 35 c and 1.00 atm is stored in a 2.85 l flask.

what is the density of the gas?

what is the molar mass of the gas?

Answers: 2

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Answers

Answer from: rafexRaex2914

1.35 g/L

Explanation:

1) Use PV = nRT to determine moles of gas present in gas:

(1.10 atm) (2.55 L) = (n) (0.08206) (328 K)

n = 0.104 mol

2) Get molar mass of gas using calculated moles:

0.104 mol x M g/mol = 3.45g

M = 3.45 g / 0.104 mol = 33.17 g/mol

Density = Molar Mass x P / RT

= 33.17 g/mol x 1.10 atm / 0.0821 L mol / atm .K x 328 K

= 1.35 g /L

Answer from: Dracopaul03

The density of the gas is 1.42g/L and the molar mass is 32.92 g/mol

Explanation:

Mass of the unknown gas=m= 4.05g

Temperature=T= 35C

Pressure exerted upon the unknown gas= P= 1 atm

Volume of the flask=V= 2.85 L

Density= ?

Density= $\frac{mass of the unknown gas}{ volume of the container containing the gas}$

= $\frac{m}{V}$

= $\frac{4.05}{2.85}$

=1.421= 1.42g/L

Molar mass of the gas=M=?

Now,

As the formula is defined

$\frac{n}{V}=\frac{P}{RT}$

n= $\frac{VP}{RT}$

(R= 0.082 L atm /K mol)

Since Temperature= 35C= (35+273) K

=308 K

n= $\frac{2.85\times 1}{0.082\times 308}$

n= $\frac{2.85}{25.26}$

n=0.1129 moles

Molar mass= M= $\frac{m}{n}$

= $\frac{4.05}{0.123}$

=32.92 g/mol

Hence the density of the gas is 1.42 g/L and the molar mass is 32.92 g/mol

Answer from: Quest

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form( $n_2$ ) .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

Answer from: Quest

c. electrical charges of ionic bonds.

i hoped this , can i get brainliest plzgod bless you

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A4.05 sample of an unknown gas at 35 c and 1.00 atm is stored in a 2.85 l flask.

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