# Please hep urrgent exam question

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#1
5V4+ + MnO4– + 8H+ 5V5+ + Mn2+ + 4H2O
What volume of 0.020 mol dm–3 KMnO4 solution is required to oxidise completely a solution containing 0.010 mol of vanadium(IV) ions? WHY IS THE ANSWER 100CM3???????
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2 years ago
#2
Molar ratio is 5:1

0.01 / 5 = 0.002 moles of MnO4

So then v=n/c --> 0.002/0.02 = 0.1dm3 --> 100cm3.
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2 years ago
#3
5V4+ + MnO4– + 8H+ 5V5+ + Mn2+ + 4H2O
What volume of 0.020 mol dm–3 KMnO4 solution is required to oxidise completely a solution containing 0.010 mol of vanadium(IV) ions? WHY IS THE ANSWER 100CM3???????
Hint: The mole ratio of V4+ to MnO4- is 5:1. And you have been giving the moles of V4+
(Don't forget to convert the cm3 to dm3 when using the formula n=cv)
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2 years ago
#4
(Original post by thekidwhogames)
Molar ratio is 5:1

0.01 / 5 = 0.002 moles of MnO4

So then v=n/c --> 0.002/0.02 = 0.1dm3 --> 100cm3.
We aren't allowed to give complete solutions on TSR
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2 years ago
#5
(Original post by Rohit_Rocks10)
We aren't allowed to give complete solutions on TSR
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#6
(Original post by thekidwhogames)
Molar ratio is 5:1

0.01 / 5 = 0.002 moles of MnO4

So then v=n/c --> 0.002/0.02 = 0.1dm3 --> 100cm3.
thank u so much, can u tell me why u divide by 5 if thats not the mr of V
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#7
(Original post by Rohit_Rocks10)
We aren't allowed to give complete solutions on TSR
i never knew that, lol since when?
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2 years ago
#8
thank u so much, can u tell me why u divide by 5 if thats not the mr of V
Well, you were given that there were 0.010 moles of Vanadium ions and we know the ratio is 5:1 so the number of moles of the other must be 0.010/5 i.e. 0.002 moles. You are also given the concentration so the formula n=cv can be rearranged to v=n/c = 0.002/0.02 = 0.1 dm^3. So we know the volume in cm3 must be 0.1 x 10^3 = 100cm3
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#9
thank u so much, can u tell me why u divide by 5 if thats not the mr of V
nevermind i get it now, thanks anyway
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2 years ago
#10
nevermind i get it now, thanks anyway
No problem.
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#11
(Original post by thekidwhogames)
No problem.
do u mind helping me with another question, i have the answer but since its a multiple choice question, theres no method to show how u get the answer
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2 years ago
#12
do u mind helping me with another question, i have the answer but since its a multiple choice question, theres no method to show how u get the answer
I don't mind - go ahead.
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#13
(Original post by thekidwhogames)
I don't mind - go ahead.
thank u!
.The removal of silicon dioxide with limestone in the Blast Furnace can be represented by the following equation. CaCO3(s) + SiO2(s) → CaSiO3(l) + CO2(g) The minimum mass of calcium carbonate needed to remove 1.00 tonne (1000 kg) of silicon dioxide is A 0.46 tonne B 0.60 tonne C 1.67 tonne D 2.18 tonne
why is it C 1.67tonne? do u first calculate moles of SiO2?
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2 years ago
#14
thank u!
.The removal of silicon dioxide with limestone in the Blast Furnace can be represented by the following equation. CaCO3(s) + SiO2(s) → CaSiO3(l) + CO2(g) The minimum mass of calcium carbonate needed to remove 1.00 tonne (1000 kg) of silicon dioxide is A 0.46 tonne B 0.60 tonne C 1.67 tonne D 2.18 tonne
why is it C 1.67tonne? do u first calculate moles of SiO2?
First note the molar ratio is 1:1.

Let's find number of moles of SiO2:

Find the Mr first --> 60. So moles = mass / Mr = 1,000,000 (grams) / 60 = 50,000/3.

So moles of CaCO3 = 50,000/3 too. We know mass =moles x Mr. Mr of CaCO3 = 40 + 12 + 3(16) = 100 so mass is 100 x 50,000/3 =1,666,666.66... grams which is roughly 1.67 tonnes (divide by 1000 to get into kg and then divide by 1000 to get tonnes).
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#15
(Original post by thekidwhogames)
First note the molar ratio is 1:1.

Let's find number of moles of SiO2:

Find the Mr first --> 60. So moles = mass / Mr = 1,000,000 (grams) / 60 = 50,000/3.

So moles of CaCO3 = 50,000/3 too. We know mass =moles x Mr. Mr of CaCO3 = 40 + 12 + 3(16) = 1000 so mass is 100 x 50,000/3 =1,666,666.66... grams which is roughly 1.67 tonnes (divide by 1000 to get into kg and then divide by 1000 to get tonnes).
thank u, what i did was find the moles of SiO2 1000000/60.1= 16638.93511 but i didnt use CaCO3, am i supposed to in this case? i just converted it back to the tonnes afterwards
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2 years ago
#16
thank u, what i did was find the moles of SiO2 1000000/60.1= 16638.93511 but i didnt use CaCO3, am i supposed to in this case? i just converted it back to the tonnes afterwards
Yes, because that would be wrong. Be careful. You know they have the same number of moles so that's the only thing in common. You have to find the mass of CaCO3 so you need to do moles x Mr so the moles (which you got from SiO2) x Mr (which from the periodic table = 40 + 12 + 3(48)). Then that's your mass - divide by a million (to go from grams to kilograms).

What you did was find the moles of SiO2 - this isn't answering the question (to find mass of CaCO3).
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2 years ago
#17
what is this question i havent seen anything like this in my life
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#18
(Original post by thekidwhogames)
Yes, because that would be wrong. Be careful. You know they have the same number of moles so that's the only thing in common. You have to find the mass of CaCO3 so you need to do moles x Mr so the moles (which you got from SiO2) x Mr (which from the periodic table = 40 + 12 + 3(48)). Then that's your mass - divide by a million (to go from grams to kilograms).

What you did was find the moles of SiO2 - this isn't answering the question (to find mass of CaCO3).
thats true, thank u very much for ur help, i really appreciate u taking ur time to explain the answers!
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2 years ago
#19
thats true, thank u very much for ur help, i really appreciate u taking ur time to explain the answers!
No problem! Any more help - let me know!

Also, some resources for A level Chemistry (what my brother and I use):

Videos:

http://scienceabove.com/ (has worksheets/ quizzess too)

Questions:

http://scienceabove.com/

https://www.scisheets.co.uk/

http://www.physicsandmathstutor.com
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#20
(Original post by thekidwhogames)
No problem! Any more help - let me know!

Also, some resources for A level Chemistry (what my brother and I use):

Videos:

http://scienceabove.com/ (has worksheets/ quizzess too)

Questions:

http://scienceabove.com/

https://www.scisheets.co.uk/

http://www.physicsandmathstutor.com
aw thank u, u are too kind. heres another question A naturally occuring sample of the element boron has a relative atomic mass of 10.8. In this sample, boron exists as two isotopes, 10B and 11B.
Calculate the percentage abundance of 10B.(the answer is 20% but not sure why)
i know the formula for this but i dont know how to use it when they dont give all the information. do u have to use x and y, 10x+11x/100=10.8?
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