Quick math and chem question

[spoog74]

Hi folks !


Quick question;

For this problem;

3x^3=33x^2

Solve for x

I got 11, but the answer says it was x=11 AND x = 0. I see how it can go both ways... But how would one know to subtract 33x^2 and equate it to zero, as opposed to just solving for x but dividing x^2 and then solving..


Another question;


Moles of AgIo3 with ksp constant? Help?
How many moles of AgIO3 ( ksp= 3.1 x 10E-8) will dissolve in 1 liter of 10 E-5 M solution of NaIO3? How would you do this?

How would you solve this?

Also i just ran in to this question;

Which of the following is not equal to .000032?

This is from test10 destroyer 2011

A. 32x10^3
B. .032 x 10^-3
C. .32 x 19^-4
D. 3200X10^-8
E. 320X10^-9

Isnt A AND E the answers? Destroyer says its E .

---

Members don't see this ad.

 

[GenieG]

For the first one, I think it's easier to just remember in cases without other numbers, x can =0 and look out for it.

For the second one, i hope this is right and think someone should check this entire process because it's been over a year since gen chem

I learned using "ICE" charts
....................................Ag...... IO3
Initial concentration ...... 0 .........1*10^-5 (from the solution)
change .........................S .........S (since they are equimolar, both are the same)
Equilibrium ...................S .........1*10^-5+S

Assume S is much smaller than 1*10^-5 (since Ksp is small) and 1*10^-5+S = ~1.5*10^-5

Ksp=[Ag][IO3] = S*(1x10^-5)=3.1*10^-8, solve for S 3.1*10^-3

 

[herkulease]

for the math destoryer question.

its an error.

just know how to move the decimals and you'll be fine. It was in the 2010 edition too. Both are valid choices. Not sure what he had in mine for A other than

 

[PhansterZ]

For the first question, bring both values to one side and set other side to 0.

3x^3 = 33x^2
3x^3 - 33x^2 = 0
3x(x-11) = 0

Set 3x = 0
x = 0

Set x-11= 0
x = 11

 

[dentalstudent2021]

For the first question, bring both values to one side and set other side to 0.

3x^3 = 33x^2
3x^3 - 33x^2 = 0
3x(x-11) = 0

Set 3x = 0
x = 0

Set x-11= 0
x = 11

so..why is this done this way ? why not x^3 / x ^2 = 33/3 ? Thanks guys, this one is killing me....(and i know this is a old post )

 

[Roy Williams]

so..why is this done this way ? why not x^3 / x ^2 = 33/3 ? Thanks guys, this one is killing me....(and i know this is a old post )

If it's a polynomial above degree 1 then you have to think about it like you would if it were a quadratic equation. If you have the equation x^2 + x = 0, you can't just make it x^2 = -x and then divide by x to get a root of -1. Yes that is one of the roots, but quadratics have 2 roots (the other is 0). Tertiary polynomials have three roots (although in the OP one is imaginary). You won't have to know how to do any crazy factoring of higher degree polynomials. If they give you one you'll be able to take out a common multiple or something. Make sense?

 

[dentalstudent2021]

If it's a polynomial above degree 1 then you have to think about it like you would if it were a quadratic equation. If you have the equation x^2 + x = 0, you can't just make it x^2 = -x and then divide by x to get a root of -1. Yes that is one of the roots, but quadratics have 2 roots (the other is 0). Tertiary polynomials have three roots (although in the OP one is imaginary). You won't have to know how to do any crazy factoring of higher degree polynomials. If they give you one you'll be able to take out a common multiple or something. Make sense?

Yes it does!! ...hope I remember it though...

 

[gangazi]

Hello
When you divide x^2 by x, you get x. While dividing, you took out one of the possible values of x... So you need to find out the value you took out

 

[dentalstudent2021]

Hello
When you divide x^2 by x, you get x. While dividing, you took out one of the possible values of x... So you need to find out the value you took out

Ahh, I see. Thanks