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Answer to Question #89813 in General Chemistry for Bri

Question #89813
40x10^5 grams of Na2CO3= How many particles of Na2CO3
Expert's answer

@$n(Na2CO3)=\dfrac{40*10^5g}{106g/mol}=37735.85moles @$

@$N(Na2CO3)=37735.85moles*6.02*10^{23}=2.272*10^{28}(particles)@$

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