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Chemistry 23 Online
OpenStudy (kittiwitti1):

What is the oxidation state of Cr2 in K2Cr2O7?

OpenStudy (kittiwitti1):

\(\huge{Cr_{2}}\) in \(\huge{K_{2}Cr_{2}O_{7}}\)

OpenStudy (kittiwitti1):

points I noted: 1. there is no charge so net oxidation must be zero, 0 2. charge of K is -1, chromium has NO charge, oxygen is TYPICALLY -2 charge.

OpenStudy (cuanchi):

+6

OpenStudy (cuanchi):

1x2 +2x +(-2) x7 =0

OpenStudy (kittiwitti1):

I did 7(-2) for Oxygen and 2(1) for Potassium

OpenStudy (kittiwitti1):

and the \(\large{x}\) is the Chromium?

OpenStudy (cuanchi):

14-2=12 12/2=6

OpenStudy (cuanchi):

1* 2 +2x +(-2) * 7 =0

OpenStudy (kittiwitti1):

I got -12 charge. To counter is +12 and Cr is 2 so 12/2 is 6 Okay I got it

OpenStudy (kittiwitti1):

How about this one?\[\large{Fe_{3}O_{4}}\]

OpenStudy (cuanchi):

8/3

OpenStudy (kittiwitti1):

I think this: net is 0 O is 4(-2) ... yes that lol

OpenStudy (kittiwitti1):

okay I have an organic compound question -- how do I know when a compound is \(\Large{propanol}\)?

OpenStudy (kittiwitti1):

actually scratch that

OpenStudy (cuanchi):

has 3 C and one OH

OpenStudy (kittiwitti1):

how do I know when a compound is \(\Large{ethanoic}\) acid

OpenStudy (kittiwitti1):

sorry I figured out the propanol one, thanks though

OpenStudy (cuanchi):

CH₃COOH (acetic acid)

OpenStudy (kittiwitti1):

yes but how do I know to get that formula? is there any tips/tricks

OpenStudy (kittiwitti1):

also... Fe in \(\huge{Fe_{3}O_{4}}\) has a decimal oxidation...?

OpenStudy (cuanchi):

yes it is an average 2 atoms has oxidation of 3 and one oxidation of 2

OpenStudy (kittiwitti1):

so Fe has \(\approx+2\) oxidation

OpenStudy (cuanchi):

+2 and +3 depends with what they react, this it is a particular compound that has both https://en.wikipedia.org/wiki/Iron(II,III)_oxide

OpenStudy (cuanchi):

OpenStudy (kittiwitti1):

I saw .gif and expected an animation.... lol

OpenStudy (kittiwitti1):

But thank you :)

OpenStudy (kittiwitti1):

!!! I KNEW THE FORMULA I WROTE WAS BAD

OpenStudy (cuanchi):

gtg see you later

OpenStudy (kittiwitti1):

alright thank you :)

OpenStudy (kittiwitti1):

see ya~

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