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Catwalker

Chemistry II- This week: Gas Volume Problem

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Okay, so I have to take Gen Chem 1 and 2 for the PA program I'm hoping to get into. I know we have some smart people here, and my professor is less than willing (not unwilling, just not as willing as I'd like) to check some of the problems I want to do for practice. Anyone want to help me with this? I'm looking for two things: correct process and correct answer. If you have to ask where I got some of the numbers, that's fine, but only ask if it's not obvious. For instance, in the problem below, I use the figure 55.85. That's the g/mol number for Fe. If you had to ask me that without knowing where to find the answer first, or at least knowing what that means....

 

I'm always going to try to work the problem myself before asking for help. If I think I might be wrong, I might post it here. I will never post a problem and ask for answers. That would not help me learn.

 

We'll start with an easy one. This one is so easy, it can barely be called Chemistry, but merely basic algebra. But, there will be harder ones. Trust me.

 

--

 

An iron bar weighed 664g. After the bar had been standing in moist air for a month, exactly one-eighth of the iron turned to rust (Fe2 O3). Calculate the final mass of the iron bar and rust.

 

Fe2O3 has 159.7g/mol.

 

Fe2 makes up (2x55.85)/159.7 = 70% of Fe2O3

 

O3 makes up (3x16.00)/159.7 = 30% of Fe2O3

 

If 7/8 of the bar remains, then 581g remains and 83g of the Fe was converted to Fe2O3.

 

By weight, there is 2.33 times more Fe2 than O3 in Fe2O3, so the amount of O3 present in the Fe2O3 is 35.57g.

 

The mass of Fe2O3 is therefore 118.57g. Add to the remaining 581g of remaining Fe and the total mass of the bar plus rust is 699.57g.

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Okay, so I have to take Gen Chem 1 and 2 for the PA program I'm hoping to get into. I know we have some smart people here, and my professor is less than willing (not unwilling, just not as willing as I'd like) to check some of the problems I want to do for practice. Anyone want to help me with this? I'm looking for two things: correct process and correct answer. If you have to ask where I got some of the numbers, that's fine, but only ask if it's not obvious. For instance, in the problem below, I use the figure 55.85. That's the g/mol number for Fe. If you had to ask me that without knowing where to find the answer first, or at least knowing what that means....

 

I'm always going to try to work the problem myself before asking for help. If I think I might be wrong, I might post it here. I will never post a problem and ask for answers. That would not help me learn.

 

We'll start with an easy one. This one is so easy, it can barely be called Chemistry, but merely basic algebra. But, there will be harder ones. Trust me.

 

--

 

An iron bar weighed 664g. After the bar had been standing in moist air for a month, exactly one-eighth of the iron turned to rust (Fe2 O3). Calculate the final mass of the iron bar and rust.

 

Fe2O3 has 159.7g/mol.

 

Fe2 makes up (2x55.85)/159.7 = 70% of Fe2O3

 

O3 makes up (3x16.00)/159.7 = 30% of Fe2O3

 

If 7/8 of the bar remains, then 581g remains and 83g of the Fe was converted to Fe2O3.

 

By weight, there is 2.33 times more Fe2 than O3 in Fe2O3, so the amount of O3 present in the Fe2O3 is 35.57g.

 

The mass of Fe2O3 is therefore 118.57g. Add to the remaining 581g of remaining Fe and the total mass of the bar plus rust is 699.57g.

 

same numbers i got.

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Like I said, it was an easy one.

 

I know there will be harder ones, though. I think I'll be working through some practice-on-my-own problems in the next couple of days, so I'll be posting them here.

 

I work through the examples in the book and if I don't get them on the first or second try, I do the practice problems for more practice. The problem I posted here was one of the problems I had to submit for a grade.

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Chemistry :wub: my second favorite subject, next to Math.

 

The information in my head might be a bit rusty but I'll help if I can.

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Yeah, I started off as a Chem major before I realized that I hated organic, and really only liked Chemistry because of the Math part. I'll try to remember to check in on this thread.

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Formula Mass (I think)

 

Formaldehyde has the formula CH2O. How many molecules are there in 0.11 g of formaldehyde?

a. 6.1 x 10–27

b. 3.7 x 10–3

c. 4

d. 2.2 x 1021

e. 6.6 x 1022

 

.11g x 1/30.03 = .0037 (moles in .11 g)

 

.0037 x (6.022 x 10^23) = 2.21 x 10^21

 

Answer D

 

Correct?

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Calculate the mass of FeS formed when 9.42 g of Fe reacts with 8.50 g of S.

 

Fe(s) + S(s) FeS(s)

a. 17.9 g

b. 87.9 g

c. 26.0 g

d. 14.8 g

e. 1.91 x 10–3 g

 

Okay, I feel silly for posting this one. It's obviously A- 17.9 grams, right? It doesn't say "reacts completely" so I'm not sure if that's the case.

 

The math:

 

9.42 g FE = .17mol

8.5g S = .27 mol

 

FeS has 87.92g/mol

 

None of the math I do comes up with any of the answers above. I can use process of elimination to eliminate B, C, E. The only two left are A and D. If the two elements don't react completely, then D should be it, but my Chemistry-fu is weak (obvious) and I can't figure out the direction to take.

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Since you form one FeS from one Fe and one S, there is a one-to-one relationship between the amount of both of the reactants and the result.

 

You have .17 moles of Fe, which is enough to form .17 moles of FeS.

 

You have .27 moles of S, which is enough to form .27 moles of FeS.

 

You will be limited by the reactant that you have the least amount of. Since you only have enough Fe to produce .17 moles of FeS, you will use up all your Fe and have .1 moles of S left over. Therefore, you have .17 * 87.91 = 14.9g FeS.

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been a long time. what's the process for determining how many atoms it takes to make a stable molecule ? is it the column in which the element resides in the periodic table tells you how many electrons are missing from their 'shell'? i forget what the count of electrons per shell (orbitals?) it is... and you still have the issue of ions, right?

 

also how do you determine what can act as a catalyst in reaction between two elements? i know intro chem usually sticks to just heat reactions for simplicity's sake...

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The oxidation number of Cl in ClO3– is ________.

a. –1

b. +7

c. +5

d. +3

e. None of the above.

 

I would say A, because (I think) the oxidation number of Cl is -1. Our book has a chart that says this (and six other charges for Cl), but otherwise, I have no idea.

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The oxidation number of Cl in ClO3– is ________.

a. –1

b. +7

c. +5

d. +3

e. None of the above.

 

I would say A, because (I think) the oxidation number of Cl is -1. Our book has a chart that says this (and six other charges for Cl), but otherwise, I have no idea.

 

The sum of compound in a molecule is = to the charge. If its neutral, IE H20, its sum is 0. In this case, its sum is negative, -1. Oxygen is -2 in most molecules. So 3(-2) = -6. So X+-6 = -1 , Cl = 5.

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The sum of compound in a molecule is = to the charge. If its neutral, IE H20, its sum is 0. In this case, its sum is negative, -1. Oxygen is -2 in most molecules. So 3(-2) = -6. So X+-6 = -1 , Cl = 5.

 

I had to read that a couple of times to get it, but I got it now. Thank you. :)

 

Here's another one:

 

What is the oxidizing agent in the following chemical reaction?

 

5H2O2 + 2MnO4– + 6H+ 2Mn2+ + 8H2O + 5O2

a. H2O2

b. MnO4–

c. H+

d. Mn2+

e. O2

 

I say B, MnO4 because an oxidizing agent is always reduced. In that equation, the Manganese Oxide(?) is reduced to just Manganese and the remainders.

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I had to read that a couple of times to get it, but I got it now. Thank you. :)

 

Here's another one:

 

What is the oxidizing agent in the following chemical reaction?

 

5H2O2 + 2MnO4– + 6H+ 2Mn2+ + 8H2O + 5O2

a. H2O2

b. MnO4–

c. H+

d. Mn2+

e. O2

 

I say B, MnO4 because an oxidizing agent is always reduced. In that equation, the Manganese Oxide(?) is reduced to just Manganese and the remainders.

 

You forget the reaction somewhere, or are you supposed to solve for the reaction first?

 

I assume you mean 5H2O2 + 2MnO4- +6H-> 2Mn2+ +8H20 +5O2

 

MnO4- is the oxidizing agent because it gains electrons, MnO4- -> Mn2+ (reduction).

 

EDIT: I skipped some steps, that you probably need while learning this.

 

Find the oxidation number of all this stuff. MnO4- = -1. O4 = -8, so Mn = +7. +7 -> +2 = -5, gain of electrons AKA reduction.

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First, I think your equation should probably be

 

5H2O2 + 2MnO4- + 6H+ --> 2Mn2+ + 8H20 + 502

 

(you were missing the arrow, made it a little hard to read)

 

The oxidizing agent is the reactant that gains electrons (aka is reduced). In this case, on the LHS, Mn has an oxidation state of 7+, and on the right 2+. In general, MnO4- is a pretty common oxidizer.

 

Also, MnO4- is called manganate.

 

Blah, Carny beat me to it, but I already had all this typed out.

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been a long time. what's the process for determining how many atoms it takes to make a stable molecule ? is it the column in which the element resides in the periodic table tells you how many electrons are missing from their 'shell'? i forget what the count of electrons per shell (orbitals?) it is... and you still have the issue of ions, right?

 

also how do you determine what can act as a catalyst in reaction between two elements? i know intro chem usually sticks to just heat reactions for simplicity's sake...

 

You asking this stuff just to clutter up this thread? You can google all of this info.

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been a long time. what's the process for determining how many atoms it takes to make a stable molecule ? is it the column in which the element resides in the periodic table tells you how many electrons are missing from their 'shell'? i forget what the count of electrons per shell (orbitals?) it is... and you still have the issue of ions, right

 

In general, you can tell what kind of ions an element will form from what column it is in, which tells you how many electrons are in which shells when you have a stable atom. In the first row, you are dealing with just the first shell, which has just one s orbital, containing at most 2 electrons. Stable Hydrogen has 1 electron in its shell, and a H+ ion (essentially a proton) has lost the electron and has an empty shell. In the second and third rows you move up to s and p orbitals, then in the fourth row you hit the transition metals which have a d orbital. Elements in the first column produce 1+ ions, the second produce 2+, etc.

 

If you want more detail I could type out more, but its been a while and I'm kind of just flying off the top of my head atm.

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The placement of the arrow was correct.

 

One last one, and from the reading I've been doing, I can't figure out how to go backward to get the answer.

 

During a titration the following data were collected: A 10 mL portion of an unknown acid solution was titrated with 1.0 M NaOH. 40 mL of the base were required to neutralize the sample. What is the molarity of the acid solution?

a. 1.0 M

b. 8.0 M

c. 4.0 M

d. 10.0 M

e. 40.0 M

 

You don't have to give me the answer, but at least get me started.

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When you are talking about molarity of an acid, you are usually talking about the Hydrogen ion concentration. To figure out how many H+ ions there were in the 10 mL of acid, you need to figure out how many OH- ions it took to fully react the acid (since one H+ reacts with one OH-).

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I am soooo tired right now. I did not get the answer.

 

This is my fault for waiting so long to work on these. I waited to the last minute. Did some Monday, yesterday and tonight. I should have spaced them out a bit more.

 

Anyway, Got an 80 on this one. So, I missed five. I passed. That's all I care about. I'll know which ones I missed in a few days.

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