Kagi98JP

24 minutes ago, Sensei said:

Volume of acetic acid which is in liquid/solid state not gaseous state..

Of course. You're absolutely right. Thanks so much! 

Hello everyone! I have been tasked with creating a airbag from a plastic sandwich bag, acetic acid, and sodium bicarbonate. 

The volume of my sandwich bag is 0.87 L, the pressure in my area is 100.6 kPa, and the temperature in the room is 295.15 K. First, I am tasked to find the mass of NaHCO₃ needed. My calculations are as follows: 

n = PV / RT 

n = (100.6 kPa)(0.87 L) / (8.314 kPa x L / mol x K)(295.15 K) 

n = 0.03566682292 mol NaHCO₃

M = 75.028 g/mol

m = n x M

m = 2.676 = 2.7 g

Now, I need to calculate the volume of acetic acid, but I run into a roadblock: 

n_HC2H3O2 = (0.03566682292 mol NaHCO₃)(1 mol HC₂H₃O₂ / 1 mol  NaHCO₃)

n_HC2H3O2 = 0.03566682292 mol

V = nRT / P

V = (0.03566682292 mol)(8.314 kPa x L / mol x K)(295.15 K) / 100.6 kPa

V = 0.8666 = 0.87 L

Huh? I doubt I need to fill the bag completely full with acetic acid to produce a reaction. What have I done wrong here? Thank you so much for the help! 

Hello again, everyone! Today, I present you all with a question about dilution and it's formula of C₁V₁=C₂V₂. I believe it's also common to see "M" in place of "C". 

My question is, "How many mL is needed to dilute a 100 mL solution of NaOH with a concentration of 10M to a concentration of 2M."

I believe the answer is 20 mL: 

C₁V₁=C₂V₂

C₁ = 10M (initial concentration) 

V₁ = ? (this is the amount of solution that must be transferred, right?)

C₂ = 2M (desired concentration)

V₂ = 100 mL (not entirely sure why this value is 100 mL, but I think it is because that's how much I want to end up with?)

(10M)(V₁) = (2M)(0.1 L)

V₁ = [(2M)(0.1 L)] / 10M

V₁ = 0.02 L = 20 mL

Now, I believe this means that 20 mL of NaOH should be diluted in 80 mL of water? This will give the NaOH a concentration of 2M and a final volume of 100 mL. Do I understand this correctly? Thanks a lot for the help  

24 minutes ago, exchemist said:

Well, actually, checking homework is not my absolute favourite pastime.......

This all looks fine to me. You are not doing a reaction with an excess of one of the reactants here, so don't worry about that. I assume from your comments at the end that 2.68 metric tonnes of coal is burnt per person per year in your country. I assume this is the US, as you have had to convert to the old Imperial units at the end - something we gave up in Britain when I was at school in the 1970s.

And yes you have one mole of carbon reacting with one mole of diatomic oxygen molecules to make one mole of CO2.

My only comment on your working is that you may find it better, soon, to start getting used to standard form (exponential or scientific notation) to express the numbers, to get rid of all these zeros which can become easy to miscount. So 2680kg = 2.68 x 10³ kg = 2.68 x 10⁶ g. But maybe you have not come to that yet.

You seem to know what you are doing so have confidence in your other calculations. I expect you have got them right.

 

Thank you so much! I'll definitely make the change over to scientific notation. I am just so hell-bent on visualizing everything that it'll take a bit of practice. Have a great day! 

C + O2 →  CO2

mC = 2680 kg = 2680000 g

mO2 =?

mCO2 = ?

MC = 12.01 g/mol

MO2 = 2MO

MO2 = (2)(16.00 g/mol)

MO2 = 32.00 g/mol

MCO2 = MC + 2MO

MCO2 = (12.01 g/mol) + (2)(16.00 g/mol)

MCO2 = 44.01 g/mol

nC = mC / MC

nC = (2680000 g) / (12.01 g/mol)

nC = 223147.3772 mol  

nO2 = (223147.3772 mol C) (1 mol O2 / 1 mol C)

nO2 = 223147.3772 mol 

nCO2 =  (223147.3772 mol O2) (1 mol CO2 / 1 mol O2)

mCO2 = nCO2 X MCO2 

mCO2 = (223147.3772 mol)(44.01 g/mol)

mCO2 = 9820716 g 

mCO2 = (9820716 g) / (454 g/lb)

mCO2 = 21632 lbs

Therefore, the mass of CO2 produced from coal is 21632 lbs every year per person. 

m = mass 

M = molar mass

n = number of moles

I am just a little thrown off by the fact that there isn't a limiting or excess reactant in this question. Every example I've done in class and in homework has included one. Since there isn't a limiting reactant, can I use the n-value of either carbon or oxygen to determine the n-value for carbon dioxide (since they're the same values)? I have 2 more questions like this that need to be checked if you guys are bored, let me know if that interests you!  

This is the exact question: 

I'm fairly certain 6.02 x 10^23 mol^-1 (Avogadro's constant) is a conversion factor. 

3 minutes ago, studiot said:

Conversion of what to what ?

The molar mass of Lithium is 6.941 grammes per mole.

Do you understand what this means ?

I understand what it means. I just don't understand whether or not 6.941 g/mol is a conversion factor. The question is worded abhorrently. 

For example, 6.02 x 10^23 mol^-1 would be a conversion factor, but is 6.941 g/mol one? You use the molar mass of Li (g/mol) to determine either the mass (g) or amount of moles (m), right? So is that an example of a conversion factor? A value, paired with 2 units (in this case g/mol) that can determine the value of another unit? 

Hey everyone! Super quick question. 

"Is 6.94 g/mol Li a conversion factor?" 

I'm just unsure of what the answer could be. I assume it is because you use the molar mass to determine the mass of a substance? Thanks for the help! 

Hello, everyone! I have a quick homework question that I assume has some relation to the reactivity series? It goes like this: "Would it be a good idea to store a solution of zinc nitrate in a container made of iron?  Why or why not?" I suspect it would be a fine idea to store the solution of zinc nitrate in a container made of iron because iron is less reactive than zinc and cannot displace it? I'm not entirely sure, as we skimmed over the reactivity series in class, so I would really appreciate a more in-depth and thorough answer. Thank you guys so much!