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Masaken

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how would you go about solving this? not sure what to do with either of these questions, was completely lost when i saw these
thanks in advance, answers are: 11 - B, 19 - D
 

carrotsss

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For question 19 I believe the answer is wrong, I did that paper and I had no idea where I went wrong so I sent it to my friend who is godly at chem and even he had no idea.
 

Masaken

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For question 19 I believe the answer is wrong, I did that paper and I had no idea where I went wrong so I sent it to my friend who is godly at chem and even he had no idea.
What would the answer be? I have no idea how to get either of these qns ;-;
 

carrotsss

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What would the answer be? I have no idea how to get either of these qns ;-;
For question 19:
- Find the number of moles of barium hydroxide
- Find the concentration of barium hydroxide in solution A using the moles found and c=n/v
- Then use c1v1=c2v2 to find v2 (the answer)
(You should get B)
 

someth1ng

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Q11. It's a sealed reaction vessel so none of the details matter apart from the total mass of the starting materials.
m(Hg) = 10 mL * 13.546 (g/mL)
m(O2) = 150 mL * 0.001429 (g/mL)

m(total) = m(Hg) + m(O2) = 135.67 g (i.e. B)
 

someth1ng

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Q19.

M(Ba(OH)2) = 171.23 g/mol
n(Ba(OH)2) = (17.1 g)/(171.23 g/mol) = 0.100 mol
c(Ba(OH)2) = 0.100 mol/0.100 L = 1.00 M

c1v1=c2v2
1.00 M * 0.00500 L = 0.04 M * v2
v2 = 0.125 L = 125 mL

As a sanity check, you can see that from 1.00 M to 0.04 M is 25x dilution. So the new volume is 5*25 mL = 125 mL.
 

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