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zbayBest ResponseYou've already chosen the best response.1
Whats the molar mass of that? I don't have a periodic table so I keep asking
 2 years ago

LukecrayonzBest ResponseYou've already chosen the best response.1
zbay, http://www.webqc.org/mmcalc.php for future use:)
 2 years ago

LukecrayonzBest ResponseYou've already chosen the best response.1
Molar mass of Na2CO3 is 105.9886 g/mol
 2 years ago

zbayBest ResponseYou've already chosen the best response.1
\[4 mole Na_2CO_3(\frac{105.99 g Na_2CO_3}{1 mole Na_2CO_3})(\frac{1 kg}{1000 g })=.424 kg Na_2CO_3\]
 2 years ago

zbayBest ResponseYou've already chosen the best response.1
Is the equation loading or does it look all crazy for you?
 2 years ago

anubhutiBest ResponseYou've already chosen the best response.2
let me first tell u how to calculate molar mass ok? u need to find of Na2CO3 so add the molar masses of the elements together multiplying them with their respective subscripts so here it goes like this Na= 23 C=12 O=16 so Na2CO3=2(23)+12+3(16)=46+12+48=96
 2 years ago

anubhutiBest ResponseYou've already chosen the best response.2
now 1 mole of sodium carbonate has 96 grams so 4 moles will have 4x106=424 oh yeah sry
 2 years ago

anubhutiBest ResponseYou've already chosen the best response.2
oops sry now that means one mole of Na2CO3 will have 106 grams
 2 years ago

XishemBest ResponseYou've already chosen the best response.0
To get the final answer, you need to convert to kg (:.
 2 years ago

anubhutiBest ResponseYou've already chosen the best response.2
well that is easy enough... from here the question can b done by the asker himself yhe main question has been solved isnt it? even so the answer is .106 kg ok?
 2 years ago

XishemBest ResponseYou've already chosen the best response.0
Well, zbay has already answered the question. It's .424kg of Na_2CO_3
 2 years ago

anubhutiBest ResponseYou've already chosen the best response.2
hah i again forgot to multiply by 4 wat a shame yeah k the answer is .424 kg i m really sry
 2 years ago

XishemBest ResponseYou've already chosen the best response.0
Don't be sorry! We're all learning here.
 2 years ago
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