omg help! chemistry help!!11

[cpt_azad]

oh god oh god oh god...need help !!11

Determine the volume of .200M NaOH required to completely neutralize 250.0mL of 0.220M H3PO4.

i need the steps on how to do this (above)

oh god


Determine the new concentration after these dilutions:

a) 150mL of .450M HCl to 4.50L

b) 40.0mL of 1.8M NaOH to 800.0mL



Calculate the concentration of each ion after mixing these solutions wher NO REACTION occurs:

a) 2.0L of 0.60M Copper (II) Nitrate and 2.0L of 0.20M Iron (III) Nitrate

b) 400.0mL of 0.200M Aluminum Chloride and 200.0mL of 0.300M Sodium Chloride

please guys, you're my last hope, i have no idea how to do this and it's due tomorrow. i forgot my binder at school and all my notes are in there, i looked on google but i can't find a good example(s) of it or steps on how to do it. PLEASE HELP ME!!!!

[S!X]

Damn meng, Im pretty sure none of these mofos in the lounge know wtf this stuff is.

[cpt_azad]

i think i figured out number 1, not sure, could someone confirm on it?:

Equation (please lemme know if its balanced properly!):

H3PO4 + 3NaOH --> 3H2O + Na3PO4

n=CV --> 0.250L(0.200M H3PO4) = 0.0550 moles H3PO4
NaOH:H3PO4 --> 3:1 --> 3(0.0550moles H3PO4) = 0.165 moles NaOH
0.165mole NaOH/0.200M --> 0.825L of NaOH is required to completely neutralize the H3PO4.

[Mr. Mulder]

yeah......yeah that looks pretty good to me :ermm:

[MCHeshPants420]

Looks completely wrong to me, unfortunately I'm off to my interview now. Maybe laters.

[cpt_azad]

damnit mchesh :angry: now ur making me even more worried ahhhhhhhhh

[99%]

Damn meng, Im pretty sure none of these mofos in the lounge know wtf this stuff is.

:cry:

[cpt_azad]

well he is rite, when was the last time anyone in the lounge actually helped someone instead of spamming the fuck out of a thread? maybe i posted this in the wrong place....

[Guillaume]

Yeah, maybe we need a "I-need-someone-else-to-do-my-homework-world". :dry:

[99%]

Yeah, maybe we need a "I-need-someone-else-to-do-my-homework-world". :dry:

I think its great when people ask for solutions to homework whether its essays/maths whatever - reminds me to get of things i once knew - if i know the answer ill say so
they generally add different topics to the board instead of usual sftwr/hrdwr/fileshr/igottagreatlink/myanusitches threads.

I never did physics nor chemistry in school - pity - tons of usefull info there for the later life.

[manker]

well he is rite, when was the last time anyone in the lounge actually helped someone instead of spamming the fuck out of a thread? maybe i posted this in the wrong place....I think you did post this in the wrong place. Most folk I talk to think you're an arse.

Have you thought about starting afresh under a new handle and not being such a prick. I mean, who are you to condescend to a group of people and for you, of all people, to berate them for spamming in the same sentence is pushing the realms of credibility a bit, dontcha think.

[MCHeshPants420]

Yeah, maybe we need a "I-need-someone-else-to-do-my-homework-world". :dry:

I think its great when people ask for solutions to homework whether its essays/maths whatever - reminds me to get of things i once knew - if i know the answer ill say so
they generally add different topics to the board instead of usual sftwr/hrdwr/fileshr/igottagreatlink/myanusitches threads.

I never did physics nor chemistry in school - pity - tons of usefull info there for the later life.


Yeah, but it's more fun to take the piss and wind them up with the wrong answers. The best thing you can say to them is to manage their time better and do their homework when they get it or at the very least start it.

[Snee]

well he is rite, when was the last time anyone in the lounge actually helped someone instead of spamming the fuck out of a thread? maybe i posted this in the wrong place....
:lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol: *




*at you, not with you.

[JPaul]

A pattern is emerging.

A h4r5h pattern, but a pattern none the less.

[DanB]

well he is rite, when was the last time anyone in the lounge actually helped someone instead of spamming the fuck out of a thread? maybe i posted this in the wrong place....


Says the spamming cawksucker himself :dry:

[Mr JP Fugley]

well he is rite, when was the last time anyone in the lounge actually helped someone instead of spamming the fuck out of a thread? maybe i posted this in the wrong place....


Says the spamming cawksucker himself :dry:
Shhh, she'll snap you like a dry twig.

[Mr JP Fugley]

Sorry wrong forum

[enoughfakefiles]

The answer is spam. :D

[Gripper]

LSD=Aciiid :)

No I have no idea about your home work,so I thought I'd :spamfrown

[HCT]

Dilutions

When given a solution, you can diluted by adding more of the solvent. to calculate how much solvent to add, use the formula: M1V1=M2V2, where M1 is the initial concentration and V1 is the initial volume. M2 is the concentration you want it diluted to and V2 is the volume of the new solution.


hope it helps:)

[exo_body]

hey man, the first one you did was wrong sorry. the equation is:
NaOH + H3PO4 -> NaH2PO4 + H20
now using the details of H3PO4
conc: 0.2 0.22
vol: 250/1000
moles: =0.22*(250/1000) (therefore, as its a 1:1 reaction, theres the same number of moles of NaOH)

now to work out the volume, you use the equation volume = moles/volume

so [0.22*(250/1000)] 0.2
will give you 0.275dm^-3, then just multiply by 1000 to get the answer in cm^-3

hope thats any help, i think its right. do you want the others done? they follow the same kinda pattern, but they seem slightly easier. well question 2 does.

[cpt_azad]

fuck you all, especially manker :D

[exo_body]

the answer to the next question (with the part a) and b) ) is:

a : 0.015 M
b: 0.09 M

again, these are the answers im getting just. if you want the working out just ask.

[JPaul]

fuck you all, especially manker :D
Break his arm with a punch captain asda.

It's the only sensible coarse.

[manker]

fuck you all, especially manker :DEven the ones who tried to help.

See, this is why my arse evaluation wasn't contradicted.

[JPaul]

fuck you all, especially manker :DEven the ones who tried to help.

Even the wrong ones :blink:

[manker]

Even the ones who tried to help.

Even the wrong ones :blink:
:unsure:

[loadedet]

All i would do would pour measured amounts of a neutraliser to the solution, then test the pH with a universal indicator then rite down the amount that was used on a table 4 even more quantafiable results use a colouromiter

[enoughfakefiles]

All i would do would pour measured amounts of a neutraliser to the solution, then test the pH with a universal indicator then rite down the amount that was used on a table 4 even more quantafiable results use a colouromiter

Are you not better off just usinig a continuum tramfunctioner. :unsure:

[cpt_azad]

hey man, the first one you did was wrong sorry. the equation is:
NaOH + H3PO4 -> NaH2PO4 + H20
now using the details of H3PO4
conc: 0.2 0.22
vol: 250/1000
moles: =0.22*(250/1000) (therefore, as its a 1:1 reaction, theres the same number of moles of NaOH)

now to work out the volume, you use the equation volume = moles/volume

so [0.22*(250/1000)] 0.2
will give you 0.275dm^-3, then just multiply by 1000 to get the answer in cm^-3

hope thats any help, i think its right. do you want the others done? they follow the same kinda pattern, but they seem slightly easier. well question 2 does.

thanks dude, i took my test today and i figured it out, i didnt' realize someone actually posted the answer, once again thanks for taking the time to help a guy out i owe you one. at least theres some good ppl in this forum.

@manker, jesus man just STFU will ya?

[MCHeshPants420]

hey man, the first one you did was wrong sorry. the equation is:
NaOH + H3PO4 -> NaH2PO4 + H20
now using the details of H3PO4
conc: 0.2 0.22
vol: 250/1000
moles: =0.22*(250/1000) (therefore, as its a 1:1 reaction, theres the same number of moles of NaOH)

now to work out the volume, you use the equation volume = moles/volume

so [0.22*(250/1000)] 0.2
will give you 0.275dm^-3, then just multiply by 1000 to get the answer in cm^-3

hope thats any help, i think its right. do you want the others done? they follow the same kinda pattern, but they seem slightly easier. well question 2 does.

I did chemistry at school so I can say this with confidence (although I'll admit I totally failed), those answers are right.