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Chemistry stuff

Fizban140Fizban140 Registered User, __BANNED USERS
edited May 2011 in Help / Advice Forum
Edit, need help again.

I can't figure these out, its multiple choice so I know I am not getting the correct answer.





4)20 mL of 3.00 M solution with 30 mL of a 5.40 M solutionb you get ?

I couldnt get an answer for this, you find the moles add that then divide by total volume? I am not sure. Either way my anwer wasn't listed.

5) 3.00 L of 3.01 M solution NaCl called solution A. 2.00 L of 2.00 M AgNO3 called solution B., mix together make solution C. Calculate conecntration in M of Na+ ions in C.

I tried this one a lot and couldnt get an answer. Its 1:1 I think so I just need to find the combined molarity but I wasn't too sure how to do that, no examples in the book.

6) Solution 1 100 ML of 3 M solution
Solution 2 50 mL 3 M
solution 3 10 mL of 3 M

All taken from 500 mL flask containing 3 M all of this is sodium carbonate

What colume in mL of water must evaporate from solution 1 in order to have concentration of 4.62 M?

7) Which of followin is not a conjugate acid base pair?
I really hate these.

A) H2SO4, SO4 2-
b) HNO3, NO3 -
C) HC2H3O2, CH3O2-
D) H2PO4 -, HPO4 2-
E) Hbr, Br-

I knowi ts going to be a change of H- which to me looks like E, D, C, B and A but maybe not A because it is 2H?

8) Calculate the pH of .044 M HCl solution

fff I did this one so many times and didn't get the right answer, none of these is an option so it could be that I guess.

Fizban140 on
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Posts

  • Raif SeveranceRaif Severance Registered User regular
    edited May 2011
    Are you sure they gave you the volume in liters and not something like deciliters?

  • Fizban140Fizban140 Registered User, __BANNED USERS
    edited May 2011
    Are you sure they gave you the volume in liters and not something like deciliters?

    Yes, the other problems used mL so I made sure, this is as written in the book is 12.5 g KNO3; 1.15 L

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  • SeGaTaiSeGaTai Registered User regular
    edited May 2011
    Most likely the solution guide is wrong since it is off by an order of magnitude-this happens more often than you would expect-your setup is correct

    PSN SeGaTai
  • Fizban140Fizban140 Registered User, __BANNED USERS
    edited May 2011
    That is good to know. Is there any sort of online calculator I could run a problem like this through to check my answers on? All the ones I have seen cost money.

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  • TheSuperWootTheSuperWoot Registered User regular
    edited May 2011
    I think your best bet would just be finding other people to check your homework with. But yeah, I've found tons of mistakes in my physics and analytical chem textbooks so if you've triple checked your units/everything else and you're just off by an order of magnitude I'd just ignore it and move on.

  • Fizban140Fizban140 Registered User, __BANNED USERS
    edited May 2011
    I am trying to study for my final, turns out I lost the answer key to a test and I can't figure out these problems and the book doesn't give specific examples for these ones.

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  • Raif SeveranceRaif Severance Registered User regular
    edited May 2011
    Fizban140 wrote: »
    Edit, need help again.

    I can't figure these out, its multiple choice so I know I am not getting the correct answer.

    1)

    3 mol N 5.66 mol H

    N^2 + 2H^2 ----> 2Nh^3 find the moles of NH^3

    I get 6 mol NH3

    Your equation isn't balanced. It should be N2 + 3H2 --> 2NH3

  • Bliss 101Bliss 101 Registered User regular
    edited May 2011
    4)20 mL of 3.00 M solution with 30 mL of a 5.40 M solution you get ?

    I couldnt get an answer for this, you find the moles add that then divide by total volume? I am not sure. Either way my anwer wasn't listed.

    If the volumes being mixed were equal, you could just calculate the average concentration from the two solutions. In this case you need to account for the fact that there's more of one solution than the other. 40% of your final 50 ml solution is 3 M, the other 60% is 5.40 M, so: (0.4 * 3M) + (0.6 * 5.40M) = your answer.

    Keep in mind that molarity is a measure of concentration. You'd do the exact same calculation if you were mixing, say, solutions of 30% and 54% ethanol. Obviously the final concentration has to be somewhere between the concentrations of the original solutions; if it isn't, you'll know you've got it wrong.

    That's the simplest way to calculate these. There are more complicated (and possibly more "correct" from a teacher's standpoint) ways to calculate this. For example, remembering that molarity (M) stands for moles/litre, you can do as you suggested yourself: find the total amount of moles, divide by final volume. In this case: (0.020 l * 3.0 mol/l + 0.030 l * 5.40 mol/l) / (0.020 l + 0.030 l). Just keep an eye on the units.

    5) 3.00 L of 3.01 M solution NaCl called solution A. 2.00 L of 2.00 M AgNO3 called solution B., mix together make solution C. Calculate conecntration in M of Na+ ions in C.

    I tried this one a lot and couldnt get an answer. Its 1:1 I think so I just need to find the combined molarity but I wasn't too sure how to do that, no examples in the book.

    Note that they only ask for the concentration of Na+ ions. Unless I'm missing something, the silver nitrate is there just to confuse you. There is no Na in AgNO3 so solution B might just as well be water for the purposes of this exercise. You're simply diluting 3 liters of NaCl solution into a final volume of 5 l.
    6) Solution 1 100 ML of 3 M solution
    Solution 2 50 mL 3 M
    solution 3 10 mL of 3 M

    All taken from 500 mL flask containing 3 M all of this is sodium carbonate

    What colume in mL of water must evaporate from solution 1 in order to have concentration of 4.62 M?

    When diluting and concentrating solutions, remember that the total amount of moles in the solution stay the same, only concentration (c) and volume (v) change. So you start with the formula c1v1 = c2v2 and calculate the unknown term. In this case you need to find out final volume: v2 = (c1v1)/c2 = (3M * 0.1l)/4.62M. This gives you the volume of the 4.62 M solution. Then just subtract that from the original volume to see how much needs to evaporate.
    7) Which of followin is not a conjugate acid base pair?
    I really hate these.

    A) H2SO4, SO4 2-
    b) HNO3, NO3 -
    C) HC2H3O2, CH3O2-
    D) H2PO4 -, HPO4 2-
    E) Hbr, Br-

    I knowi ts going to be a change of H- which to me looks like E, D, C, B and A but maybe not A because it is 2H?

    You're right with your doubts about A. Sulfuric acid dissociates in two steps (H2SO4 -> HSO4- -> SO4 2-). The conjugate base of H2SO4 is HSO4-.

    Take another look at C and see if you missed something there. =)
    8) Calculate the pH of .044 M HCl solution

    fff I did this one so many times and didn't get the right answer, none of these is an option so it could be that I guess.

    Just remember that HCl is a strong acid: all of it dissociates in solution. You ignore the H+ concentration of water, because it's many orders of magnitude below what we're looking at here. So the concentration of H+ is considered the same as the concentration of HCl. Remember that pH is just the negative 10-base logarithm of the H+ concentration, so you can calculate this simply as -log(0.044).

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