HEY

im really ill so cant go and see my teacher this week for help for the 2009 exam next wed so would be grateful if i could get loads of help off here!!

2003 MULTIPLE CHOICE

Q5 a mix of sodium chloride and sodium sulphate contains 0.6 moles of chloride ions and 0.2 mol of sulphate ions. how many sodium ions are there?

the answer is 1......??????????????????

Q14 the no. of moles in 1 mol of copper(11) phosphate is.....?? answers is 5........

( solutions say Cu3(PO4)2....i dont get why the 2 on the copper has switched with the 3- on the PO4???

Q 35 when a certain aquesous solution is diluted its conductivity decreases but ph remains constant. the solution could be.....

ethanoic acid, sodium chloride, sodium hydroxide, nitric acid??....is it just knowlegde that the answer is sodium chloride?

Q36 the ion electron equations for a redox reaction are

2I- goes to I2 + 2e-

MnO4- + 8H+ + 5E- goes to Mn(2+) + 4H2O

how many moles of iodide ions are oxidised by 1 mole of permanganate ions...answer 5.........i understand to do redox eqn but i am stuck from there

Please help, would really appreciate it as worrying bout not being well enough for exam as i have tonsilitis!!!

thanks

**0**

# 2003 mc

Started by sammy w, May 28 2009 02:03 PM

3 replies to this topic

### #1

Posted 28 May 2009 - 02:03 PM

### #2

Posted 30 May 2009 - 12:21 PM

Hi Sammy W. To avoid confusion, I'm a Chemistry Teacher and have helped out a few years in the past. I'll look back regularly between now and the exam on Wednesday and try and help as much as I can.

Then it is simple ratios.

If there are 0.2 mol of SO

If there are 0.6 mol of Cl ions then there are another 0.6 mol of Na ions.

Total Na ions = 0.4 + 0.6 = 1.0 mol

Common type of question that appears most years.

Copper (II) means Cu

2I

MnO

You have to balance the electrons lost and gained. The first equation would need to be

10I

whilst the second equation is

2MnO

This tells you that 2 mol of MnO

so

**Q5**is really a test of your formula writing. You must be able to work out that sodium sulphate is Na_{2}SO_{4}while sodium chloride is NaCl.Then it is simple ratios.

If there are 0.2 mol of SO

_{4}ions then there must be 0.4 mol of Na.If there are 0.6 mol of Cl ions then there are another 0.6 mol of Na ions.

Total Na ions = 0.4 + 0.6 = 1.0 mol

Common type of question that appears most years.

**Q14**Looks like formula writing is a weakness.Copper (II) means Cu

^{2+}while phosphate is PO_{4}^{3-}(Data Book). To balance the overall charges, 3 x Cu^{2+}and 2 x PO_{4}^{3-}.**Q35**The fact that the pH remains constant rules out anything acidic (ethanoic and nitric acid) and anything basic (sodium hydroxide) leaving just neutral sodium chloride.**Q36**As you've said2I

^{-}goes to I_{2}+**2e**^{-}MnO

_{4}^{-}+ 8H^{+}+**5e**goes to Mn^{-}^{2+}+ 4H_{2}OYou have to balance the electrons lost and gained. The first equation would need to be

**multiplied by 5**so10I

^{-}goes to 5I_{2}+**10e**^{-}whilst the second equation is

**multiplied by 2**2MnO

_{4}^{-}+ 16H^{+}+**10e**goes to 2Mn^{-}^{2+}+ 8H_{2}OThis tells you that 2 mol of MnO

_{4}^{-}reacts with 10 mol of I^{-}so

**1 mol of MnO**_{4}^{-}reacts with 5 mol of I^{-}### #3

Posted 01 June 2009 - 09:17 PM

QUOTE (ginneswatson @ May 30 2009, 01:21 PM) <{POST_SNAPBACK}>

Hi Sammy W. To avoid confusion, I'm a Chemistry Teacher and have helped out a few years in the past. I'll look back regularly between now and the exam on Wednesday and try and help as much as I can.

Then it is simple ratios.

If there are 0.2 mol of SO

If there are 0.6 mol of Cl ions then there are another 0.6 mol of Na ions.

Total Na ions = 0.4 + 0.6 = 1.0 mol

Common type of question that appears most years.

Copper (II) means Cu

2I

MnO

You have to balance the electrons lost and gained. The first equation would need to be

10I

whilst the second equation is

2MnO

This tells you that 2 mol of MnO

so

**Q5**is really a test of your formula writing. You must be able to work out that sodium sulphate is Na_{2}SO_{4}while sodium chloride is NaCl.Then it is simple ratios.

If there are 0.2 mol of SO

_{4}ions then there must be 0.4 mol of Na.If there are 0.6 mol of Cl ions then there are another 0.6 mol of Na ions.

Total Na ions = 0.4 + 0.6 = 1.0 mol

Common type of question that appears most years.

**Q14**Looks like formula writing is a weakness.Copper (II) means Cu

^{2+}while phosphate is PO_{4}^{3-}(Data Book). To balance the overall charges, 3 x Cu^{2+}and 2 x PO_{4}^{3-}.**Q35**The fact that the pH remains constant rules out anything acidic (ethanoic and nitric acid) and anything basic (sodium hydroxide) leaving just neutral sodium chloride.**Q36**As you've said2I

^{-}goes to I_{2}+**2e**^{-}MnO

_{4}^{-}+ 8H^{+}+**5e**goes to Mn^{-}^{2+}+ 4H_{2}OYou have to balance the electrons lost and gained. The first equation would need to be

**multiplied by 5**so10I

^{-}goes to 5I_{2}+**10e**^{-}whilst the second equation is

**multiplied by 2**2MnO

_{4}^{-}+ 16H^{+}+**10e**goes to 2Mn^{-}^{2+}+ 8H_{2}OThis tells you that 2 mol of MnO

_{4}^{-}reacts with 10 mol of I^{-}so

**1 mol of MnO**_{4}^{-}reacts with 5 mol of I^{-}thank you sooo much. I wish i had looked at the website sooner but i thought noone was going to help!

I won't make it into school for last minute revision either so hopefully you wil be able to help with any last minute quesries tomorrow!

thanks again

x

### #4

Posted 02 June 2009 - 05:53 AM

I'm free for 7 periods today and will have a reasonable number of my Higher & Advanced Higher pupils in my room revising. I will, however, keep an eye on HSN and should be able to respond to questions.

Hope you're feeling better.

Hope you're feeling better.

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