# NEED URGENT IGCSE Chemistry Help!!!!

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I am taking a exam in two weeks.

I need help on these three questions:

I would also like it if someone explained to me in a really simple form.

1: 16g sample of an iron oxide contains 84g of iron and 32g of oxygen. Calculate the Emperial Formula?

I UNDERSTAND What I have to do

Divide 84 and 56( 1.5 mole) then 32 and 16( 2 moles) but what after that?

Yes I know that you multiply both by 3 and get Fe304 BUT How do you know to multiply by three to get Fe304, I need that explained to me

2: So calculate the Molar Enthalpy change for burning of Propanol (C3H17OH)

Yes i've got that it's 60

Now I looked up the answer and it's 18.8 x 60

Can someone please explain where you get 18.8 from? That's all i'm stuck on for this question.

3:

2KOH + H2SO4 > k2SO4 + 2H20

question:

25.0cm of KOH was used and it's concentration was 0.200mol

Calculate the amount of KOH used in moles

Calculate the amount of H2SO4 used in moles

Calculate the concentration pf the acid in solution in mol/dm3

Calculate the concentration of the KOH Solution in g/dm3

^ Need this explained to be also..

I need help on these three questions:

I would also like it if someone explained to me in a really simple form.

1: 16g sample of an iron oxide contains 84g of iron and 32g of oxygen. Calculate the Emperial Formula?

I UNDERSTAND What I have to do

Divide 84 and 56( 1.5 mole) then 32 and 16( 2 moles) but what after that?

Yes I know that you multiply both by 3 and get Fe304 BUT How do you know to multiply by three to get Fe304, I need that explained to me

2: So calculate the Molar Enthalpy change for burning of Propanol (C3H17OH)

Yes i've got that it's 60

Now I looked up the answer and it's 18.8 x 60

Can someone please explain where you get 18.8 from? That's all i'm stuck on for this question.

3:

2KOH + H2SO4 > k2SO4 + 2H20

question:

25.0cm of KOH was used and it's concentration was 0.200mol

Calculate the amount of KOH used in moles

Calculate the amount of H2SO4 used in moles

Calculate the concentration pf the acid in solution in mol/dm3

Calculate the concentration of the KOH Solution in g/dm3

^ Need this explained to be also..

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#4

(Original post by

I am taking a exam in two weeks.

I need help on these three questions:

I would also like it if someone explained to me in a really simple form.

1: 16g sample of an iron oxide contains 84g of iron and 32g of oxygen. Calculate the Emperial Formula?

I UNDERSTAND What I have to do

Divide 84 and 56( 1.5 mole) then 32 and 16( 2 moles) but what after that?

Yes I know that you multiply both by 3 and get Fe304 BUT How do you know to multiply by three to get Fe304, I need that explained to me

2: So calculate the Molar Enthalpy change for burning of Propanol (C3H17OH)

Yes i've got that it's 60

Now I looked up the answer and it's 18.8 x 60

Can someone please explain where you get 18.8 from? That's all i'm stuck on for this question.

3:

2KOH + H2SO4 > k2SO4 + 2H20

question:

25.0cm of KOH was used and it's concentration was 0.200mol

Calculate the amount of KOH used in moles

Calculate the amount of H2SO4 used in moles

Calculate the concentration pf the acid in solution in mol/dm3

Calculate the concentration of the KOH Solution in g/dm3

^ Need this explained to be also..

**elmosandy**)I am taking a exam in two weeks.

I need help on these three questions:

I would also like it if someone explained to me in a really simple form.

1: 16g sample of an iron oxide contains 84g of iron and 32g of oxygen. Calculate the Emperial Formula?

I UNDERSTAND What I have to do

Divide 84 and 56( 1.5 mole) then 32 and 16( 2 moles) but what after that?

Yes I know that you multiply both by 3 and get Fe304 BUT How do you know to multiply by three to get Fe304, I need that explained to me

2: So calculate the Molar Enthalpy change for burning of Propanol (C3H17OH)

Yes i've got that it's 60

Now I looked up the answer and it's 18.8 x 60

Can someone please explain where you get 18.8 from? That's all i'm stuck on for this question.

3:

2KOH + H2SO4 > k2SO4 + 2H20

question:

25.0cm of KOH was used and it's concentration was 0.200mol

Calculate the amount of KOH used in moles

Calculate the amount of H2SO4 used in moles

Calculate the concentration pf the acid in solution in mol/dm3

Calculate the concentration of the KOH Solution in g/dm3

^ Need this explained to be also..

For the first question what you have to do is:

Divide the Relative Atomic Mass of Iron by the actual amount you have so 84/56 = 1.5, and the same for oxygen 32/16 = 2.

The ratio above (1.5) isn't a whole number so you times both numbers by 2, so the ratio is 1.5*2, 2*2, so amount of iron to oxygen is 3 to 4.

So the empirical formula is Fe3O4.

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(Original post by

I am doing that exam too.

For the first question what you have to do is:

Divide the Relative Atomic Mass of Iron by the actual amount you have so 84/56 = 1.5, and the same for oxygen 32/16 = 2.

The ratio above (1.5) isn't a whole number so you times both numbers by 2, so the ratio is 1.5*2, 2*2, so amount of iron to oxygen is 3 to 4.

So the empirical formula is Fe3O4.

**IGCSEKid**)I am doing that exam too.

For the first question what you have to do is:

Divide the Relative Atomic Mass of Iron by the actual amount you have so 84/56 = 1.5, and the same for oxygen 32/16 = 2.

The ratio above (1.5) isn't a whole number so you times both numbers by 2, so the ratio is 1.5*2, 2*2, so amount of iron to oxygen is 3 to 4.

So the empirical formula is Fe3O4.

I am doing Edexcel IGCSE Chemistry, is that what you're doing?

And can you help me with the two other questions please? x

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#8

(Original post by

Thank you so much! omg taken losts of stress away!.

I am doing Edexcel IGCSE Chemistry, is that what you're doing?

And can you help me with the two other questions please? x

**elmosandy**)Thank you so much! omg taken losts of stress away!.

I am doing Edexcel IGCSE Chemistry, is that what you're doing?

And can you help me with the two other questions please? x

_{2}SO

_{4}à K

_{2}SO

_{4}+ H

_{2}O

Information we know:

25.0 cm

^{3}of 2KOH used, with concentration of 0.2 mol/dm

^{3}

i) Calculate the amount of KOH used in moles

ii) Calculate the amount of H

_{2}SO

_{4}used in moles

iii) Calculate the concentration of the acid in solution in mol/dm

^{3}

iv) Calculate the concentration of the KOH solution in g/dm

^{3}

Specification points we need to know and use:

1.27

1.19

i) We know these formulae:

Concentration (in mol/dm

^{3}) = Number of moles / Volume (in dm

^{3})

This can rearranged to ‘Number of moles = Concentration (in mol/dm

^{3}) x Volume (in dm

^{3})’

Volume of KOH = 25.0 cm

^{3}

= 0.025 dm

^{3}(We divide by 1000 to convert from cm

^{3}to dm

^{3})

Concentration of KOH = 0.2 mol/dm

^{3}

Moles of KOH = 0.005 moles

ii) If you look at the chemical equation the mole ratio of KOH to H

_{2}SO

_{4 }is 2:1 because the numbers in front of formulae represent moles.

Therefore to find the number of moles of H

_{2}SO

_{4}you just divide your previous answer by 2.

0.005 moles / 2 = 0.0025 moles

iii) You can’t do this because you only know one value (number of moles) but you don’t know the volume and hence you can’t calculate concentration.

iv) You have 0.2 mol/dm

^{3}of KOH solution.

The M

_{r}of KOH is 39 + 16 + 1 = 56

You know that Moles = Mass (in g) / M

_{r}à Mass (in g) = Moles x M

_{r}

In 1 dm

^{3}, 0 .2 (I know this is the concentration but since we are talking about in 1dm

^{3}we just use this value for the number of moles) x 56

0.2 x 56 = 11.2 g/dm

^{3}

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(Original post by

Q3) 2KOH + H

Information we know:

25.0 cm

i) Calculate the amount of KOH used in moles

ii) Calculate the amount of H

iii) Calculate the concentration of the acid in solution in mol/dm

iv) Calculate the concentration of the KOH solution in g/dm

Specification points we need to know and use:

1.27

1.19

i) We know these formulae:

Concentration (in mol/dm

This can rearranged to ‘Number of moles = Concentration (in mol/dm

Volume of KOH = 25.0 cm

= 0.025 dm

Concentration of KOH = 0.2 mol/dm

Moles of KOH = 0.005 moles

ii) If you look at the chemical equation the mole ratio of KOH to H

Therefore to find the number of moles of H

0.005 moles / 2 = 0.0025 moles

iii) You can’t do this because you only know one value (number of moles) but you don’t know the volume and hence you can’t calculate concentration.

iv) You have 0.2 mol/dm

The M

You know that Moles = Mass (in g) / M

In 1 dm

0.2 x 56 = 11.2 g/dm

**Naveen1412**)Q3) 2KOH + H

_{2}SO_{4}à K_{2}SO_{4}+ H_{2}OInformation we know:

25.0 cm

^{3}of 2KOH used, with concentration of 0.2 mol/dm^{3}i) Calculate the amount of KOH used in moles

ii) Calculate the amount of H

_{2}SO_{4}used in molesiii) Calculate the concentration of the acid in solution in mol/dm

^{3}iv) Calculate the concentration of the KOH solution in g/dm

^{3}Specification points we need to know and use:

1.27

1.19

i) We know these formulae:

Concentration (in mol/dm

^{3}) = Number of moles / Volume (in dm^{3})This can rearranged to ‘Number of moles = Concentration (in mol/dm

^{3}) x Volume (in dm^{3})’Volume of KOH = 25.0 cm

^{3}= 0.025 dm

^{3}(We divide by 1000 to convert from cm^{3}to dm^{3})Concentration of KOH = 0.2 mol/dm

^{3}Moles of KOH = 0.005 moles

ii) If you look at the chemical equation the mole ratio of KOH to H

_{2}SO_{4 }is 2:1 because the numbers in front of formulae represent moles.Therefore to find the number of moles of H

_{2}SO_{4}you just divide your previous answer by 2.0.005 moles / 2 = 0.0025 moles

iii) You can’t do this because you only know one value (number of moles) but you don’t know the volume and hence you can’t calculate concentration.

iv) You have 0.2 mol/dm

^{3}of KOH solution.The M

_{r}of KOH is 39 + 16 + 1 = 56You know that Moles = Mass (in g) / M

_{r}à Mass (in g) = Moles x M_{r}In 1 dm

^{3}, 0 .2 (I know this is the concentration but since we are talking about in 1dm^{3}we just use this value for the number of moles) x 560.2 x 56 = 11.2 g/dm

^{3}I'm going to sound like a ***** but yeah I need it to be explained in the most simple form, like seriously I've asked Question 1 on Yahoo Answers five times, asked other chem questions too I could not understand any of the answers, even a IGCSE Student answered my question and I still couldn't understand, I have only got Question 1 after @IGCSEKid explained to me, so yeah explain again? ( I kind of got it but yeah got lost after ratio lol :/ )

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#12

**elmosandy**)

Thank you so much! omg taken losts of stress away!.

I am doing Edexcel IGCSE Chemistry, is that what you're doing?

And can you help me with the two other questions please? x

@Naveen1412 Is right but I'll try explain it more simpler if I can.

**For Q3 (i):**

There is a general formula for chemistry problems this is: Number of moles = Concentration (in mol/dm

^{3}) x Volume (in dm

^{3})

The volume you have is 25.0 cm

^{3}so we convert this volume to the one we have to use in the formula which is dm

^{3}to do this we divide the volume by 1000 so we now have 0.025dm

^{3}.

We also need the Concentration but this is already given in the question.

So now we can put the numbers into the formula: Number Of moles = 0.2*0.025 = 0.005 Moles

DONE!

**For Q3 (ii):**

If we look at the equation again: 2KOH + H2SO4 > k2SO4 + 2H20

The ratio for 2KOH and H2SO4 is 2 : 1

We know that 2KOH is 0.005 Moles so to get H2SO4 we just divide by 2 so: 0.0025 moles

DONE!

**For Q3 (iii):**@Naveen1412 is right you con't do this question.

**For Q3 (iv):**

I'm going to be honest with you, I don't know the answer to this question

And for the Q2,

I have never really seen this question in exam papers that I have done so I don't think that sort of question will come up.

Hopefully this helps you

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(Original post by

No problem Yes I'm doing that exam.

@Naveen1412 Is right but I'll try explain it more simpler if I can.

There is a general formula for chemistry problems this is: Number of moles = Concentration (in mol/dm

The volume you have is 25.0 cm

We also need the Concentration but this is already given in the question.

So now we can put the numbers into the formula: Number Of moles = 0.2*0.025 = 0.005 Moles

DONE!

If we look at the equation again: 2KOH + H2SO4 > k2SO4 + 2H20

The ratio for 2KOH and H2SO4 is 2 : 1

We know that 2KOH is 0.005 Moles so to get H2SO4 we just divide by 2 so: 0.0025 moles

DONE!

I'm going to be honest with you, I don't know the answer to this question

And for the Q2,

I have never really seen this question in exam papers that I have done so I don't think that sort of question will come up.

Hopefully this helps you

**IGCSEKid**)No problem Yes I'm doing that exam.

@Naveen1412 Is right but I'll try explain it more simpler if I can.

**For Q3 (i):**There is a general formula for chemistry problems this is: Number of moles = Concentration (in mol/dm

^{3}) x Volume (in dm^{3})The volume you have is 25.0 cm

^{3}so we convert this volume to the one we have to use in the formula which is dm^{3}to do this we divide the volume by 1000 so we now have 0.025dm^{3}.We also need the Concentration but this is already given in the question.

So now we can put the numbers into the formula: Number Of moles = 0.2*0.025 = 0.005 Moles

DONE!

**For Q3 (ii):**If we look at the equation again: 2KOH + H2SO4 > k2SO4 + 2H20

The ratio for 2KOH and H2SO4 is 2 : 1

We know that 2KOH is 0.005 Moles so to get H2SO4 we just divide by 2 so: 0.0025 moles

DONE!

**For Q3 (iii):**@Naveen1412 is right you con't do this question.**For Q3 (iv):**I'm going to be honest with you, I don't know the answer to this question

And for the Q2,

I have never really seen this question in exam papers that I have done so I don't think that sort of question will come up.

Hopefully this helps you

Thankyou! Hopefull Q3 iii and iv don't come up in the exam!

TYVM Both of you guys! :P

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#14

(Original post by

Thankyou! Hopefull Q3 iii and iv don't come up in the exam!

TYVM Both of you guys! :P

**elmosandy**)Thankyou! Hopefull Q3 iii and iv don't come up in the exam!

TYVM Both of you guys! :P

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#15

**elmosandy**)

Thankyou! Hopefull Q3 iii and iv don't come up in the exam!

TYVM Both of you guys! :P

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(Original post by

you can't actually do part 3 iii and for part iv if you dont understand i can send you a couple of screenshots from the cgp revision guide and the textbook which explain how to answer that type of question. Do you have either of those books?

**Naveen1412**)you can't actually do part 3 iii and for part iv if you dont understand i can send you a couple of screenshots from the cgp revision guide and the textbook which explain how to answer that type of question. Do you have either of those books?

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**Naveen1412**)

you can't actually do part 3 iii and for part iv if you dont understand i can send you a couple of screenshots from the cgp revision guide and the textbook which explain how to answer that type of question. Do you have either of those books?

One last question:

194.0g of Lead reacts with 20.g of Oxygen to form a Lead Oxide

What volume of oxygen gas, in cm3 at room tempreature and pressure weighs 20.0g?

I know rtp: is /24 but yeah confused :/

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#18

Volume (in dm3) = Moles x 24

Volume (in dm3) = (Mass/Mr) x 24

Mass of oxygen being used is 20g

Mr of oxygen (O2) is 32

Therefore input the numbers into the equation:

20/32 x 24

=15 dm3

=15 000 cm3 (to convert from dm3 to cm3 times by 1000)

Volume (in dm3) = (Mass/Mr) x 24

Mass of oxygen being used is 20g

Mr of oxygen (O2) is 32

Therefore input the numbers into the equation:

20/32 x 24

=15 dm3

=15 000 cm3 (to convert from dm3 to cm3 times by 1000)

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(Original post by

Volume (in dm3) = Moles x 24

Volume (in dm3) = (Mass/Mr) x 24

Mass of oxygen being used is 20g

Mr of oxygen (O2) is 32

Therefore input the numbers into the equation:

20/32 x 24

=15 dm3

=15 000 cm3 (to convert from dm3 to cm3 times by 1000)

**Naveen1412**)Volume (in dm3) = Moles x 24

Volume (in dm3) = (Mass/Mr) x 24

Mass of oxygen being used is 20g

Mr of oxygen (O2) is 32

Therefore input the numbers into the equation:

20/32 x 24

=15 dm3

=15 000 cm3 (to convert from dm3 to cm3 times by 1000)

And are you a GCSE Student or?:

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#20

(Original post by

Thankyou and please pm me the screenshots x

And are you a GCSE Student or?:

**elmosandy**)Thankyou and please pm me the screenshots x

And are you a GCSE Student or?:

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