Hey Guys !

I'm writing the chem 1A03 exam this Monday and I'm in a bad situation. I have a 41% in the course, I'm not enjoying chemistry at all. I'm really worried as the final exam is 35% and 35 marks. Overall, I get super stressed and nervous for exams and I'm in panic mode at the moment. Is there any hope for me to pass the course? I do not want to do anything with chemistry in the future, I actually want to do an Honors Bio with Enviro Sci.

Any advice would be greatly appreciated !

Step 1) Calm down. If you are stressed while studying you will retain less knowledge. try putting on some relaxing music (here's my suggestion Solo Piano-Sky.fm)
Step 2) Evaluate what resources you should use to study. If you made nice notes those are a prime place to start, however I do have advice if your notes are less than adequate. Go into avenue and download the lecture notes for the chapters in .pdf format. Now open them in a pdf editor, such as the freeware Foxit PDF editor or NitroPDF. You know how some parts of the slides are made 'invisible' to try and make people come to the lectures? When they posted the slides all they did to hide sections was place a white box over the important bits. You can delete this box by clicking on it and hitting delete, and now you can view all the important points you may have missed by not attending lectures.
Step 3) Tutorial questions and practice tests. Using these will help you identify what kind of problems you have difficulty with. When you have difficulty with a problem first try to solve it on your own, but if that fails feel free to post here in this thread. Many MI users, myself included, are quite good with Chem and would be more than happy to assist you.
Step 4) Get a good night's rest Sunday night, and eat full healthy meals Monday. If your body isn't in good shape how can you expect your brain to be?

Best of luck to you!

Originally Posted by BernaBee

Hey Guys !

I'm writing the chem 1A03 exam this Monday and I'm in a bad situation. I have a 41% in the course, I'm not enjoying chemistry at all. I'm really worried as the final exam is 35% and 35 marks. Overall, I get super stressed and nervous for exams and I'm in panic mode at the moment. Is there any hope for me to pass the course? I do not want to do anything with chemistry in the future, I actually want to do an Honors Bio with Enviro Sci.

Any advice would be greatly appreciated !

You have (0.41*65) = 26.65
You need (50-26.65)/35*100% = 66.7% on the final exam to pass the course.

You can't get 67%+ on the exam to pass the course?

The best way to study for it would be to do the practice/assignment problems yourself and make sure you can reproduce the answers (don't just look at the solutions and think you get it), memorize all the material you won't have access to.

Study and pray.

Btw even in biology you will have to deal with some chemistry

You're asking for help a day before? Yeah, good strategy there.

Originally Posted by britb

You're asking for help a day before? Yeah, good strategy there.

He's got time. It's 8pm on Saturday. Exam is on Monday at 12:30pm. Lots of time to pull off a 67% on the exam

Well since you guys offered

Your stomach (volume = 2.5 L) has a pH of 1.00 because of the presence of HCl. How many grams of Mg(OH)2 (58.3 g/mol) do you need to add to completely neutralize the acid in your stomach?

My answer is 7.3E-12 grams, however answer is 7.3 grams. I cant figure out 10^-12 portion.

Any help is appreciated.

Originally Posted by BernaBee

Hey Guys !

I'm writing the chem 1A03 exam this Monday and I'm in a bad situation. I have a 41% in the course, I'm not enjoying chemistry at all. I'm really worried as the final exam is 35% and 35 marks. Overall, I get super stressed and nervous for exams and I'm in panic mode at the moment. Is there any hope for me to pass the course? I do not want to do anything with chemistry in the future, I actually want to do an Honors Bio with Enviro Sci.

Any advice would be greatly appreciated !

I was also failing the course going into the exam... ended up with an 11.
Theres always hope

Good Luck- study hard... go over all the class notes, then do ALL the tutorials - and try the midterms... make sure you understand why each answer is what it is. You'll be fine. Dont Stress

again... Good Luck!

http://www.khanacademy.org/

also... check out that website, the links under chemistry are very relevant to that course. If you were like me and didn't go to class either... then those will come in handy for any concepts you didn't understand.

ps. not going to class= big mistake.

Originally Posted by pavlov

Well since you guys offered

Your stomach (volume = 2.5 L) has a pH of 1.00 because of the presence of HCl. How many grams of Mg(OH)2 (58.3 g/mol) do you need to add to completely neutralize the acid in your stomach?

My answer is 7.3E-12 grams, however answer is 7.3 grams. I cant figure out 10^-12 portion.

Any help is appreciated.

I'll include every step to try and be helpful.


Always start with a balanced equation:
Mg(OH)2 + 2HCl -> MgCl2 + 2H2O

Figure Out what you need and how you get it from the given information work from there by substitution
Mass Mg(OH)2= Moles Mg(OH)2 * Molar Mass Mg(OH)2

Mass Mg(OH)2=(Moles HCl *( 1 mole Mg(OH)2 / 2 moles HCl)) *(58.3 g/mol Mg(OH)2)

Mass Mg(OH)2=((2.5L* Molarity of HCl) *( 1 mole Mg(OH)2 / 2 moles HCl)) *(58.3 g/mol Mg(OH)2)

Mass Mg(OH)2=((2.5L* (Molarity H3O+ * 1mol HCl/H3O+) ) *( 1 mole Mg(OH)2 / 2 moles HCl)) *(58.3 g/mol Mg(OH)2)

Mass Mg(OH)2=((2.5L* (10 ^-PH * 1mol HCl/H3O+) ) *( 1 mole Mg(OH)2 / 2 moles HCl)) *(58.3 g/mol Mg(OH)2)

Mass Mg(OH)2=((2.5L* (10 ^-1 * 1mol HCl/H3O+) ) *( 1 mole Mg(OH)2 / 2 moles HCl)) *(58.3 g/mol Mg(OH)2)

Mass Mg(OH)2=(0.25mol HCl ) *( 1 mole Mg(OH)2 / 2 moles HCl)) *(58.3 g/mol Mg(OH)2)

Mass Mg(OH)2=(0.125mol Mg(OH)2 ) *(58.3 g/mol Mg(OH)2)

Mass Mg(OH)2=7.2875

Mass Mg(OH)2=7.3

Mass Mg(OH)2=See the thanks button down there? Push it.

Originally Posted by REPLEKIA/.

I'll include every step to try and be helpful.


Always start with a balanced equation:
Mg(OH)2 + 2HCl -> MgCl2 + 2H2O

Figure Out what you need and how you get it from the given information work from there by substitution
Mass Mg(OH)2= Moles Mg(OH)2 * Molar Mass Mg(OH)2

Mass Mg(OH)2=(Moles HCl *( 1 mole Mg(OH)2 / 2 moles HCl)) *(58.3 g/mol Mg(OH)2)

Mass Mg(OH)2=((2.5L* Molarity of HCl) *( 1 mole Mg(OH)2 / 2 moles HCl)) *(58.3 g/mol Mg(OH)2)

Mass Mg(OH)2=((2.5L* (Molarity H3O+ * 1mol HCl/H3O+) ) *( 1 mole Mg(OH)2 / 2 moles HCl)) *(58.3 g/mol Mg(OH)2)

Mass Mg(OH)2=((2.5L* (10 ^-PH * 1mol HCl/H3O+) ) *( 1 mole Mg(OH)2 / 2 moles HCl)) *(58.3 g/mol Mg(OH)2)

Mass Mg(OH)2=((2.5L* (10 ^-1 * 1mol HCl/H3O+) ) *( 1 mole Mg(OH)2 / 2 moles HCl)) *(58.3 g/mol Mg(OH)2)

Mass Mg(OH)2=(0.25mol HCl ) *( 1 mole Mg(OH)2 / 2 moles HCl)) *(58.3 g/mol Mg(OH)2)

Mass Mg(OH)2=(0.125mol Mg(OH)2 ) *(58.3 g/mol Mg(OH)2)

Mass Mg(OH)2=7.2875

Mass Mg(OH)2=7.3

Mass Mg(OH)2=See the thanks button down there? Push it.

Thanks! I took a completely different approach, but your method seems simpler and easier to follow.

Originally Posted by pavlov

Well since you guys offered

Your stomach (volume = 2.5 L) has a pH of 1.00 because of the presence of HCl. How many grams of Mg(OH)2 (58.3 g/mol) do you need to add to completely neutralize the acid in your stomach?

My answer is 7.3E-12 grams, however answer is 7.3 grams. I cant figure out 10^-12 portion.

Any help is appreciated.

How did you calculate it? It should be done by [H^+]=10^(-1) mol / L, then multiply that by the volume, then divide it by two because of the molar ratio between HCl and Mg(OH)2, and then multiply it by the molar mass of Mg(OH)2. I got 7.2875 g. Your mistake is most likely because of either converting litres, which you did not need to do, or getting the logarithmic formula for the pH level wrong.

NVm, you beat me to it.

Originally Posted by pavlov

Thanks! I took a completely different approach, but your method seems simpler and easier to follow.

I guess while I'm handing out advice to people:

REPLEKIA/.'s guide to "OH NO!, my answer isn't one of the multiple choice answers!":

1) Time to whip out your periodic table. I personally find that the number cause of error in a problem is an incorrect balanced equation. Make 100% sure your equation is right!

2) Check Exponents and signs. It's really easy to forget to square a set of brackets on your calculator, or maybe your - became a + when you rewrote it in the next step by accident. Check for these little math errors

3) If you still don't have an answer, do the problem again, but write out all the steps this time making sure to include units. Hopefully writing out all the steps with units will make you realize what you did wrong. With all the conversions required in Chem it's easy to mess up and use the wrong units. Example: using moles of HCl instead of Mg(OH)2 in the above problem.

4) If you still can't find an answer try and see if there are any answers available that make no sense, ex. +'ve answers when you know the answer is -'ve. cross out all nonsensical answers and guess whatever seems most likely. If you can't cross any out follow the golden rule "When in doubt, pick C. Unless 'all of the above' is a choice or it's a T/F question"

Originally Posted by justinsftw

How did you calculate it? It should be done by [H^+]=10^(-1) mol / L, then multiply that by the volume, then divide it by two because of the molar ratio between HCl and Mg(OH)2, and then multiply it by the molar mass of Mg(OH)2. I got 7.2875 g. Your mistake is most likely because of either converting litres, which you did not need to do, or getting the logarithmic formula for the pH level wrong.

NVm, you beat me to it.

I approached it similarly to you.

I converted pH to pOH found [OH-], multiply by volume. Since we have two OH- in Mg(OH)2 I divided by 2 then multipled by molar mass of Mg(OH)2.

pOH = 14 - 1 = 13 -> [OH-] = 1E-13 mol/L * 2.5 L = 2.5E-13 mol OH-
2.5E-13 mol OH- * 1 mol Mg(OH)2/2 mol OH- = 1.25E-13 mol Mg(OH)2 * 58.3 g/mol = 7.3E-12 g

However, this is incorrect so im guessing my method is invalid.

Originally Posted by pavlov

I approached it similarly to you.

I converted pH to pOH found [OH-], multiply by volume. Since we have two OH- in Mg(OH)2 I divided by 2 then multipled by molar mass of Mg(OH)2.

pOH = 14 - 1 = 13 -> [OH-] = 1E-13 mol/L * 2.5 L = 2.5E-13 mol OH-
2.5E-13 mol OH- * 1 mol Mg(OH)2/2 mol OH- = 1.25E-13 mol Mg(OH)2 * 58.3 g/mol = 7.3E-12 g

However, this is incorrect so im guessing my method is invalid.

The reason that doesn't work is because you are finding the pOH of the HCl solution. Once the Mg(OH)2 is added the pH/pOH will be 7, as this is the definition of a neutralization reaction. You could try to use a pH of 7 to try and find the answer but all it will end up telling you is that there is one mole of HCl for every 2 moles of Mg(OH)2, at which point you'd have to solve like I already did above anyway.

Although I must admit, aside from that one mistake all the concepts are applied well.

Originally Posted by REPLEKIA/.

I guess while I'm handing out advice to people:

REPLEKIA/.'s guide to "OH NO!, my answer isn't one of the multiple choice answers!":

1) Time to whip out your periodic table. I personally find that the number cause of error in a problem is an incorrect balanced equation. Make 100% sure your equation is right!

2) Check Exponents and signs. It's really easy to forget to square a set of brackets on your calculator, or maybe your - became a + when you rewrote it in the next step by accident. Check for these little math errors

3) If you still don't have an answer, do the problem again, but write out all the steps this time making sure to include units. Hopefully writing out all the steps with units will make you realize what you did wrong. With all the conversions required in Chem it's easy to mess up and use the wrong units. Example: using moles of HCl instead of Mg(OH)2 in the above problem.

4) If you still can't find an answer try and see if there are any answers available that make no sense, ex. +'ve answers when you know the answer is -'ve. cross out all nonsensical answers and guess whatever seems most likely. If you can't cross any out follow the golden rule "When in doubt, pick C. Unless 'all of the above' is a choice or it's a T/F question"

Also, when possible, cross out the stuff that is for sure wrong before attempting (this is less applicable to calculation0style questions).