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Chemistry 19 Online
OpenStudy (vera_ewing):

Question 2: Law of Conservation of Mass (3 points) The clouds of Jupiter are composed mostly of ammonia, NH3, formed by a synthesis reaction between hydrogen, H2, and nitrogen, N2. Assuming all of the substances are gaseous, express the law of conservation of mass quantitatively for this reaction by completing the particle model drawing.

OpenStudy (vera_ewing):

@Technodynamic here is the first one

OpenStudy (technodynamic):

Alright, step 1 They want you to write out the balanced equation first.

OpenStudy (vera_ewing):

Ok so what would the balanced equation be?

OpenStudy (technodynamic):

They say NH3 is formed, so when something forms what has happened? You now have a what? Synthesis reaction between H2 "and" N2", the word and means + right?

OpenStudy (vera_ewing):

Yes ok. So what is the balanced equation?

OpenStudy (technodynamic):

They want you write the balanced equation. Do you know how to write a balanced equation?

OpenStudy (vera_ewing):

No I don't know how. And I'm in a hurry because this thing is timed lol that's why I really need your help

OpenStudy (technodynamic):

First, The reactant side must equal the product side Also for this problem i think they want you to draw the ball and stick for this.

OpenStudy (vera_ewing):

Can you show me how?

OpenStudy (vera_ewing):

@Technodynamic

OpenStudy (technodynamic):

H2 + N2 => NH3 This isn't balanced. Each side must have the equal amounts of numbers. How many H's are on the left side? How many H's are on the right side?

OpenStudy (vera_ewing):

2 H's on the left and 3 H's on the right

OpenStudy (technodynamic):

Excellent! So to make them even, how many numbers will we need to make them even. Let's say they both need to equal 6. How can we can 6 H's? on both sides.

OpenStudy (vera_ewing):

Multiply by 3 on the left side right?

OpenStudy (technodynamic):

That's correct! Next, do the same for the other H.

OpenStudy (vera_ewing):

So now there are 9 H's on the right side?

OpenStudy (technodynamic):

No, it has to equal six. You have 3 x ? = 6

OpenStudy (vera_ewing):

oh 3 x 2

OpenStudy (technodynamic):

Yeah you got it!

OpenStudy (vera_ewing):

so what would the final balanced equation be?

OpenStudy (technodynamic):

Now, you see that both H's are equal, so you balanced the H's. Next, look at the N's to see if they balance.

OpenStudy (vera_ewing):

2 N's on the left and 3 on the right?

OpenStudy (technodynamic):

Correct, there are 2 on the left. On the right, no...but you are close. Write out what you have, so I can check.

OpenStudy (vera_ewing):

oh wait there is only one N on the right?

OpenStudy (technodynamic):

You're still close. You have 2 on the left, so you must have the same on the right side too so it will balance.

OpenStudy (vera_ewing):

so would i multiply the N on the right by 2?

OpenStudy (technodynamic):

Yes! That is correct

OpenStudy (vera_ewing):

Ok so what is the final balanced equation? I want to make sure I get it right

OpenStudy (technodynamic):

Show me what you wrote so far.

OpenStudy (vera_ewing):

I have H6+N2 ---> N2H6

OpenStudy (vera_ewing):

Oh ok thanks. Now what do I do for the model?

OpenStudy (technodynamic):

3H2 + N2 => 2NH3

OpenStudy (technodynamic):

Do you understand why it is balanced in the way that it is?

OpenStudy (vera_ewing):

Yes I do. So now how would I do the particle model drawing?

OpenStudy (technodynamic):

When gave you the original information for the equation, that doesn't change. You just simply multiply to equal them out. I wanted to point that out to you because I know that it will be on the test.

OpenStudy (vera_ewing):

Ok thanks! So can you show me how to do the next part of the question?

OpenStudy (technodynamic):

For the drawing, now that you have the balanced information. You must now draw it out for ball-and-stick. Which is easy b/c you have the answer already. Just draw it

OpenStudy (vera_ewing):

Can you show me what it should look like?

OpenStudy (technodynamic):

Take your molecules and draw them. Like with how many you have. Now You should be all set!

OpenStudy (vera_ewing):

Ok but I'm not really sure what it would look like..can you just show me?

OpenStudy (technodynamic):

First let me ask you, do you know what a molecule looks like?

OpenStudy (vera_ewing):

I guess so...yes.

OpenStudy (technodynamic):

Ok great! I'll get you started. |dw:1425078483018:dw|

OpenStudy (vera_ewing):

|dw:1425078574353:dw|

OpenStudy (technodynamic):

Correct for the N2 just add the stick inbetween.

OpenStudy (vera_ewing):

Do i add the stick for the H3

OpenStudy (technodynamic):

For the 2NH3 Draw a 1 circle with N attached to 3 circles of H You need two groups of these.

OpenStudy (vera_ewing):

Can you show me

OpenStudy (technodynamic):

So basically your pic is correct. You just forgot to add the stick in-between the circles. For the last H3, Put an N to it

OpenStudy (vera_ewing):

Wait are you sure?

OpenStudy (vera_ewing):

I'm so confused can you show me

OpenStudy (technodynamic):

Notice my pic has 3 pairs of H's. They have a stick in-between.

OpenStudy (technodynamic):

Put a stick in-between your N2

OpenStudy (vera_ewing):

Is your pic the completed thing for the question? Is that the answer or is there more? Can you show me the complete model? I'm not good at drawing on here

OpenStudy (technodynamic):

There's more, but I'm teaching you how to do it. The reason i'm teaching you is because you will definitely see this again lol. Trust me.

OpenStudy (technodynamic):

The pic is not completed. I did the first part to show you how, and so that you can manage the rest.

OpenStudy (technodynamic):

Just follow the first pic i sent you, and do the same for N2 and 2NH3.

OpenStudy (technodynamic):

You have N2 correct, just add the stick in there.

OpenStudy (vera_ewing):

Haha ok so is this the full model |dw:1425079073199:dw|

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