1. Imagine you have a sealed 20.0 L balloon filled with helium gas at 750 mmHg in the house at 25°C.
a. If you brought it outside on a winter day and its temperature dropped to -10°C.
Determine the volume of the balloon. Assume that the pressure in the balloon will stay
constant.

17.6510 L

Explanation:

First we should get the number of moles of helium here by Boyle's law

PV=nRT

P=750/760= 0.9868 atm

T=25+273=298 kelvin

R= 0.08206

V= 20L

so

n=PV/RT

n=0.9868×20/0.08206×298

n=0.80707 mol

Then use the same law

V=0.80707×0.08206×263/0.9869=

17.6510L

SO THE VOLUME WILL BE 17.6510 L

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.

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"