Recent content by pandamonium786

So could I do this then: H3X + 3NaOH --------> Na3X + 3H2O NaOH = 3[H3X] = 3(0.251mol/L) = 0.753mol/L H3X=48.028g/2L=24.014g/L Therefore, 0.753mol/L = 24.014g/L Molar Mass = 24.014/0.753 = 31.9g/mol

Hey guys! So I have this chemistry problem which I'm finding very difficult. QUESTION: You have an unknown triprotic acid, H3X, and have titrated it with 22.53mL of NaOH. The pH of the unknown solution is 2.04 and the concentration is 48.028g/2L. While the concentration of the NaOH is...

So would 400^40 be greater because So that means that 20^100 is actually greater because 20^ 20x20 is bigger than 400^ 20x2.

P Please expand. Also how would you reach the answer logically?

Well I flipped it because you can't have negative exponents. Also how would you solve the problem?

then what do you propose the answer is and how would you get to it?

Homework Statement Evaluate: 6^1 + 6 ^ −1 / 6^1 − 6 ^ −1 Homework Equations Exponent Laws The Attempt at a Solution 6^1 + 6 ^ −1 / 6^1 − 6 ^ −1 = 6^1 + 6 ^1 / 6^1 − 6 ^ 1 (I flipped the side - top or bottom - of the negative exponent numbers) =12/0 but i think i did it wrong

Does that mean that since 20^2 = 400 the 400^40 has to be greater than 20^100. Since 400^2 alone is approximately greater than 20^2.

well 20 x 20 = 400 or 400/20 = 20

What do you mean?

I don't know if i am right, but would there be 3 0's in 20^100 and around 4 0's in 400^40?

Thanks again :D

Homework Statement Without using a calculator and without evaluating expressions, determine which is greater. 20^100 or 400^40 Homework Equations None, however some exponent laws and possibly mental math are applicable. The Attempt at a Solution I don't know how to reach an answer without...

i understand that but could negative numbers like -1 work?

Homework Statement For which non-zero value of x is the equation -x^ -4 = (-x)^ -4 true? Explain. Homework Equations None. Other than applicable exponent laws. The Attempt at a Solution I know how to use the guess and check method. But I was wondering how to reach the answer faster and how...