chem prelim (1 Viewer)

[bubblesss]

Re: 回复: Re: 回复: Re: chem help!!!!!!1 please

You can't do it if excess wasn't there.

Be assured that they'll never give you a question where there isn't excess though.

For example - you should know about combustion right? In complete combustion between a hydrocarbon and oxygen the products are only water and CO2. However, this only takes place if there is excess of O2, if there wasn't an excess then the products would be CO and C along with CO2, we're unable to guess how much of each is produced, thus we can't really do any calculations.

thanks * hundred times*

 

[tommykins]

回复: Re: 回复: Re: 回复: Re: chem help!!!!!!1 please

I'll just add you on MSN in case you have any other queries regarding maths physics or chem, is that okay with you?

 

[what else?]

Re: 回复: Re: 回复: Re: 回复: Re: chem help!!!!!!1 please

Its fine with me. I always have questions haha

 

[tommykins]

回复: Re: 回复: Re: 回复: Re: 回复: Re: chem help!!!!!!1 please

PM me your email then.

 

[Continuum]

Re: 回复: Re: 回复: Re: chem help!!!!!!1 please

You can't do it if excess wasn't there.

Be assured that they'll never give you a question where there isn't excess though.

For example - you should know about combustion right? In complete combustion between a hydrocarbon and oxygen the products are only water and CO2. However, this only takes place if there is excess of O2, if there wasn't an excess then the products would be CO and C along with CO2, we're unable to guess how much of each is produced, thus we can't really do any calculations.

They can... It's testing your ability to work out stuff with limiting reagents...

I'll just add you on MSN in case you have any other queries regarding maths physics or chem, is that okay with you?

Pedo.

Jokes.

 

[tommykins]

回复: Re: 回复: Re: 回复: Re: chem help!!!!!!1 please

They can... It's testing your ability to work out stuff with limiting reagents...

And if they do, it'd be something like 75% of it reacts, so you still use the original equation but at 75%.

yet to see one however.

 

[bubblesss]

chem help please

3.5g of barium nitrate was mixed with 4g of ammonium sulfate solution. what mass of precipitate will form?

any help will be appreciated!!!!!!!!!!!

thanks in advance

 

[kaz1]

Re: chem help please

3.5g of barium nitrate was mixed with 4g of ammonium sulfate solution. what mass of precipitate will form?

any help will be appreciated!!!!!!!!!!!

thanks in advance

Write a balanced equation. Find how many moles of each reactants and then use the ratios to find how many moles of Barium Sulfate(the precipitant). Then convert the moles to mass and you have the answer.

 

[clintmyster]

Re: chem help please

first write your equation.

Ba(NO3)2 (aq) + (NH4)2SO4 (aq) => BaSO4(s) + 2NH4NO3(aq)

therefore BaSO4 is the precipitate (refer to solubility rules)

now we know that we have 3.5g of Ba(NO3)2 so we work out the number of moles and you get 0.013393...mols (formula used is n=m(in g)/M (Relative atomic mass i think it is called)

now its a 1:1 mol ratio with BaSO4 so same number of moles.
Now sub back into n = m/M
rearrange in terms of m and you get 2.88g

pretty sure thats right

 

[bubblesss]

Re: chem help please

ahhh can someone please do it for me because i dont get the moles part of it????????????????

 

[bubblesss]

Re: chem help please

first write your equation.

Ba(NO3)2 (aq) + (NH4)2SO4 (aq) => BaSO4(s) + 2NH4NO3(aq)

therefore BaSO4 is the precipitate (refer to solubility rules)

now we know that we have 3.5g of Ba(NO3)2 so we work out the number of moles and you get 0.013393...mols (formula used is n=m(in g)/M (Relative atomic mass i think it is called)

now its a 1:1 mol ratio with BaSO4 so same number of moles.
Now sub back into n = m/M
rearrange in terms of m and you get 2.88g

pretty sure thats right

yep i got that as well but the answer says 3.03381g???????

 

[clintmyster]

Re: chem help please

you did? i know what i did wrong. Make sure your using the right weights of elements. I mistaken sulfur for nitrogen in my second calculation but i didn exactly get that answer.

got 3.126989897g which equals 3.1g cos of two significant figures

 

[bubblesss]

more help guys!!!!!!1111

50 ml of 0.2 mol/l CuSO4 is taken from a 500 mL solution and added to a second beaker, then diluted with water to make a total of 4oomL. Calculate the concentration of the second solution?

thanks in advance!!!!!!!!!

 

[bubblesss]

uh 1 more question please!!!11

if the heat released by glucose is 147.8 J

then wat is the heat released by 1 mole?

 

[lyounamu]

more help guys!!!!!!1111

50 ml of 0.2 mol/l CuSO4 is taken from a 500 mL solution and added to a second beaker, then diluted with water to make a total of 4oomL. Calculate the concentration of the second solution?

thanks in advance!!!!!!!!!

Use the formula: c1v1 = c2v2