1. Calculate the density of oxygen, O2, under each of the following conditions: STP 1.00 atm and 20.0 ∘C2. To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 3.6-L bulb, then filled it with the gas at 1.30 atm and 20.0 ∘C and weighed it again. The difference in mass was 5.5 g. Identify the gas.3. A sample of gas has an initial volume of 31.6L at a pressure of 1.1 atm. If the sample is compressed to a volume of 10.2 L, what will its pressure be?4. Use the molar volume of a gas at STP to calculate the density (in g/L) of carbon dioxide gas at STP.

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Never gonna let you down

Never gonna run around and desert you

Never gonna make you cry

Never gonna say goodbye

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(Give you up)

(Ooh) Never gonna give, never gonna give

(Give you up)

We've known each other for so long

Your heart's been aching but you're too shy to say it

Inside we both know what's been going on

We know the game and we're gonna play it

I just wanna tell you how I'm feeling

Gotta make you understand

Never gonna give you up

Never gonna let you down

Never gonna run around and desert you

Never gonna make you cry

Never gonna say goodbye

Never gonna tell a lie and hurt you

Never gonna give you up

Never gonna let you down

Never gonna run around and desert you

Never gonna make you cry

Never gonna say goodbye

Never gonna tell a lie and hurt you

Never gonna give you up

Never gonna let you down

Never gonna run around and desert you

Never gonna make you cry​

Explanation:

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() .

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.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%