Chemistry, 25.04.2020 03:52, britneyvaughn219

A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium is added to the container and the volume adjusted so that the pressure remains the same. How many grams of helium are added to the cylinder if the volume changes from 2.00 L to 2.70 L

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answered: krystalhurst97

The mass of helium added is approximately 0.7 grams

Explanation:

Here we have

Initial mass of helium gas in the cylinder = 2.00 g

Initial pressure of gas = Final pressure of gas

Initial volume of gas = 2.00 L

Final volume of gas = 2.70 L = ‪0.0027 m

Initial temperature of the gas = room temperature = 21 °C

Molar mass of helium gas = 4.003 g/mol

Therefore, the number of moles, n of helium is given as

$n = \frac{Mass}{Molar \ mass} = \frac{2.00 \ g}{4.003 \ g/mol} = 0.49963 \ moles$

The pressure of the gas in the cylinder is given by;

$P = \frac{nRT}{V} = \frac{0.49963 \times 8.3145 \times 294.15}{0.002} = 610,969.31676 \ Pa$

Therefore, when the volume is increased to 2.7 by adding more helium, we have

$n = \frac{PV}{RT} = \frac{610,969.31676 \times 0.0027 }{8.3145 \times 294.15} = 0.67449 \ moles$

That is the number of moles of helium added is given by;

0.67449 - 0.49963 = 0.174864 moles

Mass of helium added = Number of moles added × Molar mass of helium

0.174864 moles × 4.003 g/mol = ‭0.69998111 ≈ 0.7 grams.

answered: Guest

specific gravity is the density of a substance divided by the density of water. since (at standard temperature and pressure) water has a density of 1 gram/cm3, and since all of the units cancel, specific gravity is usually very close to the same value as density

so "c" is the answer

plz give brainliest

answered: Guest

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.