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javeya
Questions (2)
A particular household ammonia solution (d= 0.97g/mL) is 6.8% NH3 by mass.
How many milliliters of this solution should be
2 answers
721 views
A particular household ammonia solution (d= 0.97g/mL) is 6.8% NH3 by mass.
How many milliliters of this solution should be
3 answers
624 views
Answers (1)
first i did was ph = 11.50 PH= 14-11.50 = 2.5 10^-2.5 = 0.0032 M Volume - 0.65 L d=0.97g/mL Mass %= 6.8% 1000mL= 1L d=m/v m=d X v m= 0.97g/mL X 1000mL m=970g mNH3 = 970g X 0.068=65.96g of NH3 MM of NH3 = 14g/mol + 3 X 1g/mol=17g/mol nNH3 = 65.96g/17g/mol
