chemistry calculation - help needed

[jedaerrow]

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hi, i'm working on problems on Ksp

and i got stuck at the part where i have to do some complicated calculation

Ksp = (4.5 x 10^-10) [3(4.5 x 10^-10)]^3 = 27x(4.5 x 10^-10)^4

how do i do the calculation with numbers complicated like that?

the answer is 1.1 x 10^-36

any suggestions ?

here's another similar one

2.1 x 10^-6 = 4x^3

i need to solve for x

i know i have to divide it by 4

and do the 3 root thingy

but how do i do it by hand?

 

[cyrano]

Can you post the entire text and answer of the original problems?

 

[Streetwolf]

hi, i'm working on problems on Ksp

and i got stuck at the part where i have to do some complicated calculation

Ksp = (4.5 x 10^-10) [3(4.5 x 10^-10)]^3 = 27x(4.5 x 10^-10)^4

how do i do the calculation with numbers complicated like that?

the answer is 1.1 x 10^-36

any suggestions ?

This one is incredibly simple actually. Notice that you have the expression (4.5 * 10^-10) there twice. One time to the 1st power, and one time to the 3rd power. So when multiplied together it is really to the 4th power (as in the answer). Then the only number left is the 3 in front of the second 4.5*10^-10. Since the 3 is also being cubed, it becomes 27.

Want to see that a bit easier? Replace all occurrences of 4.5*10^-10 with 'x' and pretend it's just an algebra problem. You'd have (x)[3(x)]^3 = x*27x^3 = 27x^4. When you plug 4.5*10^-10 back into 'x' you'll get that answer.

here's another similar one

2.1 x 10^-6 = 4x^3

i need to solve for x

i know i have to divide it by 4

and do the 3 root thingy

but how do i do it by hand?

Divide by 4 and get 0.525 * 10^-6 = x^3. This can be rewritten as 525 * 10^-9 = x^3 (WHY?).

Now take the cube root of each side. This should be estimated near 8 * 10^-3 since 8^3 = 512. So I would put down something like 8.05 * 10^-3.

Answers in bold.

 

[okie7701]

hi,
so sorry that i post another question here cuz i was hoping if anyone could help as i came across a fundamental chem calculation but i cant seem to solve it either, it goes like this :

The no. of moles of Na2CO3. 10H20 is 0.02mol, find the no. of moles of sodium in 5.72g of sodium carbonate crystals
and to find the mass of water present in 5.72g of the sodium carbonate crystals.

i don't know how to make use of the 0.02mol to solve the questions..

 

[hellonurse]

hi,
so sorry that i post another question here cuz i was hoping if anyone could help as i came across a fundamental chem calculation but i cant seem to solve it either, it goes like this :

The no. of moles of Na2CO3. 10H20 is 0.02mol, find the no. of moles of sodium in 5.72g of sodium carbonate crystals
and to find the mass of water present in 5.72g of the sodium carbonate crystals.

i don't know how to make use of the 0.02mol to solve the questions..

I don't really know what's going on here. My best GUESS would be:
where Na2CO3.10H2O=X


(5.72gX) x (mol/286gX) x (.02mol X/1 mol X) x (2 mol Na/1 mol X) = .0008

Is the answer .04 mol Na? or is it .0008?

If it's not either of those..then I really don't know.

 

[DiscoBalls]

NOw, I'm not going to say that you absolutely won't encounter a problem like this on the real DAT, since they have license to ask you pretty much whatever they want. However, others have said this before in this forum, and my experience on the actual DAT I found it to be true as well... That the actual DAT GC section has alot more conceptual problems than calculation heavy problems, and I don'r recall having to do anything quite so complex on test day.

 

[okie7701]

ic 🙂thanks alot for help and advices..