Gen Chem Questions from ADA prep material

[etoolky1]

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Someone please help! I keeping missing these questions on the practice DAT from the ADA (the application book). Can someone who understands chemistry please explain.

1. A flask weighs 95g when empty when filled with 200ml of a certain liquid, the weight is 328g. What volume in ml would 1000g of liquid occupy?
the correct ans is🙁200)(1000)/(328-95) for some reason I chose 200/(1000)(328-95)

2. Which of t he following will be the final volume when 400ml of .6M HCl is diluted to .5M HCl?
the correct ans🙁400/1)(.6/.5) I chose (1000-400)/1(.5/.6)

3. If 25ml of .5M NaOH neutralizes 35ml of a monoprotic acid, which of the following is the molarity of the acid?
a. [(0.025)(0.5)]/0.035
b. 0.035/[(0.025)(0.5)]
c. 0.025/[(0.5)(0.035)]
d. (0.025)(0.5)(0.035)
e. [(0.035)(0.5)]/0.025
the correct ans. is a and I chose d

I know these are all dilution problems and I am probably making a simple mistake. I just don't understand....Some one please help

 

[E. coli]

Someone please help! I keeping missing these questions on the practice DAT from the ADA (the application book). Can someone who understands chemistry please explain.

1. A flask weighs 95g when empty when filled with 200ml of a certain liquid, the weight is 328g. What volume in ml would 1000g of liquid occupy?
the correct ans is🙁200)(1000)/(328-95) for some reason I chose 200/(1000)(328-95)

2. Which of t he following will be the final volume when 400ml of .6M HCl is diluted to .5M HCl?
the correct ans🙁400/1)(.6/.5) I chose (1000-400)/1(.5/.6)

3. If 25ml of .5M NaOH neutralizes 35ml of a monoprotic acid, which of the following is the molarity of the acid?
a. [(0.025)(0.5)]/0.035
b. 0.035/[(0.025)(0.5)]
c. 0.025/[(0.5)(0.035)]
d. (0.025)(0.5)(0.035)
e. [(0.035)(0.5)]/0.025
the correct ans. is a and I chose d

I know these are all dilution problems and I am probably making a simple mistake. I just don't understand....Some one please help

1. Initially, you have 95 g. You add 200 mL, and the weight became 325g. So the weight of the 200 mL is 325 - 95. The density of the solution would be (325-95)g / 200 mL.
Now you have to figure out the volume that 1000 g of the solution fit. So you have divide the 1000 g by the density. (g) / (g /v) = V

2. The initial concentration is 0.6 M and the volume is 400 mL. The total number of moles (molarity * concentration) would be (400 mL * 0.6M). If you dilute this amount of moles into 0.5 M, you would need 400 mL * 0.6M/ 0.5M. (basically, the equation is initial molarity * initial volume = final molarity * final volume).

3. Neutralization reaction is when same amount of acid and base react to form water and salt. Therefore, the total moles of acid and base is the same. THe equation of moles is concentration (M) * volume (L). Therefore, it's just concentration of acid * volume of acid = concentration of base * volume of base. Therefore, the answer is 0.025 L * 0.5 M / 0.035L.

IF you still do not understand, you need to find a chemistry textbook.

 

[teefRcool]

i was doing that test do can someone explain

If 1 mole of N2 and 1 mole of H2 are mixed and allowed to react according to the equation
N2 + 3H2 --> 2NH3 what is the maximum number of moles of NH3 that coudl be produced?

answers are
a) 2/3
b) 3/2
c) 2/1
d) 1/2
e) 1/1

Could someone please explain this i'm bad at chem also.

 

[arash]

i was doing that test do can someone explain

If 1 mole of N2 and 1 mole of H2 are mixed and allowed to react according to the equation
N2 + 3H2 --> 2NH3 what is the maximum number of moles of NH3 that coudl be produced?

answers are
a) 2/3
b) 3/2
c) 2/1
d) 1/2
e) 1/1

Could someone please explain this i'm bad at chem also.

I believe the answer is A, every mole of N2 with 3 moles of H2 give you 2 moles of NH3. Now if you have 1 mole of N2 + 1 mole of H2, you will get 1/3 x 2 moles of NH3, which give you 2/3. Another way to attack this problem is by understanding the concept of limiting reagent. Here you don't have enough H2 to yield the product as the equation states. so what you can do is assume you have 1 mole of H2, and see how many moles of NH3 you can get from it. H2 is all you care about from the reactents since it is the limiting reagent:

(2 grams H2) x (1 mole H2/2 grams) x (2moles NH3/3 mole H2) x (17 grams NH3/1 mole NH3)= 11.3 which is like 2/3s of 17

 

[teefRcool]

I believe the answer is A, every mole of N2 with 3 moles of H2 give you 2 moles of NH3. Now if you have 1 mole of N2 + 1 mole of H2, you will get 1/3 x 2 moles of NH3, which give you 2/3. Another way to attack this problem is by understanding the concept of limiting reagent. Here you don't have enough H2 to yield the product as the equation states. so what you can do is assume you have 1 mole of H2, and see how many moles of NH3 you can get from it. H2 is all you care about from the reactents since it is the limiting reagent:

(2 grams H2) x (1 mole H2/2 grams) x (2moles NH3/3 mole H2) x (17 grams NH3/1 mole NH3)= 11.3 which is like 2/3s of 17

how did you figure out 1/3. what does that respresnt. tnx sorry just don't see it for some reason. thanks for the help.

 

[arash]

how did you figure out 1/3. what does that respresnt. tnx sorry just don't see it for some reason. thanks for the help.

We have:

N2 + 3H2 --> 2NH3 ,

This means you need 3 moles of H2 + 1 mole N2 to get 2 moles of NH3, but you only have 1 mole H2 (1/3 of what you need for 3 moles NH3), and since it is the limiting reagent you will get 1/3 times 2 moles of NH3. I really can't explain it better. If you still dont get it you need to roll back that chemistry text book and go to the chapter that talks about chemical reactions and stoichiometry. Good luck

 

[teefRcool]

We have:

N2 + 3H2 --> 2NH3 ,

This means you need 3 moles of H2 + 1 mole N2 to get 2 moles of NH3, but you only have 1 mole H2 (1/3 of what you need for 3 moles NH3), and since it is the limiting reagent you will get 1/3 times 2 moles of NH3. I really can't explain it better. If you still dont get it you need to roll back that chemistry text book and go to the chapter that talks about chemical reactions and stoichiometry. Good luck

I got it thanks a lot. I need to brush up a bit.