Chemistry, 04.11.2019 06:31 NutMeg6544

Ineed on this question on my chemistry final exam practice sheet !
50 points

Ineed on this question on my chemistry final exam practice sheet ! 50 points

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Answer from: emmdolzx

There's basically 3 states of matter, gas, liquid and solid. (There's also plasma but that's whole another thing)

Sample 1 should be liquid. This is because liquid takes up spaces of the container. The molecules can slip past each other at ease as they have a weaker bond to each other. However, they couldn't be compressed so the volume couldn't be changed even in different containers, their volume is fixed.

Sample 2 should be solid. It has a fixed shape and thus it stays in the shape of a sphere. The molecules are fixed to each other, they have a strong bond between them and they cannot move around, therefore, their shape cannot be changed unless you heat them up and change it to another state.

Sample 3 should be a gas. Not only they take up the whole container, they can also change its volume depending on the size of the container. Each molecule has a lot of space around and they can move freely in the container as they want to. If the container is smaller, the molecules can be compressed together tighter. We can prove this by filling a syringe with air, cover the tip with your finger, and try to push the plunger. You should be able to push the plunger without air shooting out because the molecules can be compressed.

So there you have it, we can determine the different state of matter depending on their own properties.

Answer from: djennings8055

Sample 1: Liquid

Sample 2: Solid

Sample 3: Gas

Answer from: Quest

Yeah irbyoundo this you will get the a swer wuick

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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