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Chemistry 6 Online
OpenStudy (anonymous):

What is the percent composition of sodium hydrogen carbonate (NaHCO3)? Na = 44.21%, H = 1.94%, C = 23.09%, O = 30.76% Na = 30.76%, H = 1.94%, C = 23.09%, O = 44.21% Na = 27.36%, H = 1.20%, C = 14.30%, O = 57.14% Na = 57.14%, H = 1.20%, C = 14.30%, O = 27.36%

OpenStudy (anonymous):

@iPwnBunnies

OpenStudy (anonymous):

is it 2nd row?

OpenStudy (anonymous):

@thushananth01

OpenStudy (anonymous):

@Somy

OpenStudy (somy):

waht is total Mr u found?

OpenStudy (anonymous):

idk i guessed

OpenStudy (somy):

NaHCO3 Na= 23 H= 1 C= 12 O= 16*3 so total is 84 total

OpenStudy (somy):

84 = 100% 23= X

OpenStudy (somy):

this will be % of Na do same thing for the others

OpenStudy (somy):

there is no need tho if u find this u can understand which will be ur option

OpenStudy (anonymous):

so wouldnt i get Na = 44.21%, H = 1.94%, C = 23.09%, O = 30.76%

OpenStudy (anonymous):

@Somy

OpenStudy (somy):

23*100/84=

OpenStudy (anonymous):

its the last row then

OpenStudy (anonymous):

@Somy

OpenStudy (somy):

calculate it =_=

OpenStudy (anonymous):

i tried!

OpenStudy (anonymous):

:(

OpenStudy (anonymous):

idk this one

OpenStudy (somy):

u multiplied 23*100 and then divided by 84 ?

OpenStudy (anonymous):

i got 27.38

OpenStudy (somy):

im telling u to do it in calculator

OpenStudy (somy):

here u go

OpenStudy (somy):

that's ur answer

OpenStudy (somy):

Na = 27.36%, H = 1.20%, C = 14.30%, O = 57.14%

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