J.R. S. answered 11/16/20
Ph.D. University Professor with 10+ years Tutoring Experience
a) 61 g / 329 g (x100) = 19% (m/m)
b) 3.5 g / 3.5 g + 30.5 g (x100) = 3.5 g/34 g (x100) = 10.% (m/m)
c) 50.5 g / 253 g (x100) = 20.0% (m/m)
Ailyn J.
asked 11/16/20a) 61 g of NaOH in 329 g of NaOH solution
b) 3.5 g of KOH and 30.5 g of H2O
c) 50.5 g of Na2CO3 in 253.0 g of Na2CO3 solution
J.R. S. answered 11/16/20
Ph.D. University Professor with 10+ years Tutoring Experience
a) 61 g / 329 g (x100) = 19% (m/m)
b) 3.5 g / 3.5 g + 30.5 g (x100) = 3.5 g/34 g (x100) = 10.% (m/m)
c) 50.5 g / 253 g (x100) = 20.0% (m/m)
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