Chem question! help

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livehard

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Hi, guys
Hopefully these questions are not competely embarassing. I am taking my DAT soon so please help!
#1. During a titration it was determined that 30.00m L of a 0.100 M Ce 4+ solution was required to react completely with 20.00 m L of a 0.150 M Fe2+ solution. Which one of the following reactions occurred?

Answer: Ce4+ + Fe2+ à Fe3+ + Ce3+
(I am not too sure how to approach this question!)

#2. A 49 gram sample of sulfuric acid, H2SO4 (98g/mol) contains

A. 1 mol of S atoms
B. 16 grams of O
C. 2.0 grams of H
D. 2 moles of O atoms
E. 1 mole of molecules

I know that sulfuric acid has 0.5 moles so I tried to find number of atoms by multiplying by 6.02x10^23. But I don’t see the answer. 😕
The answer is D

thanks for help!
 
OHHHHHH ! i totally misunderstood the question...stupidly...
thanx

any idea for number 1..?!
 

Hi, guys
Hopefully these questions are not competely embarassing. I am taking my DAT soon so please help!
#1. During a titration it was determined that 30.00m L of a 0.100 M Ce 4+ solution was required to react completely with 20.00 m L of a 0.150 M Fe2+ solution. Which one of the following reactions occurred?

Answer: Ce4+ + Fe2+ à Fe3+ + Ce3+
(I am not too sure how to approach this question!)

#2. A 49 gram sample of sulfuric acid, H2SO4 (98g/mol) contains

A. 1 mol of S atoms
B. 16 grams of O
C. 2.0 grams of H
D. 2 moles of O atoms
E. 1 mole of molecules

I know that sulfuric acid has 0.5 moles so I tried to find number of atoms by multiplying by 6.02x10^23. But I don’t see the answer. 😕
The answer is D

thanks for help!

for question 1 yo have to know the concept;
you know we have
(n1)M1V1=(n2)M2V2 n1 and n2 are oxidation, reduction numbers from equation for Ce+4 and Fe+2 when you plug the numbers you will have:
n1(30 X0.1)=n2(0.150 X20)===>
n1=n2 in other hand:
Fe+2=========> Fe+3 + 1e
Ce+4 + 1e========> Ce+3
n1 and n2 can be 2 or 3 the point is must be equal that is fit the answer you gave here.
maybe there is simlpe explanation somebody can help you. 🙂
 
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