You are mixing x ml of wine, which is 0.0030 m in sugar, with 20.0 ml of wine, which is 0.00010 m in sugar. the resulting wine is mixed with 30.0 ml of wine, which is 0.0020 m in sugar. the resulting wine is 0.0015 m in sugar. what is the value of x? assume that the volumes are additive.

The rule that we will use to solve this problem is:
M1V1 + M2V2 + M3V3 = MtVt
where:
M1 is the molarity of first solution = 0.003 molar
V1 is the volume of the first solution which is the X that we want to find
M2 is the molarity of second solution = 0.0001 molar
V2 is the volume of second solution = 20 ml
M3 is the molarity of third solution = 0.002 molar
V3 is the volume of the third solution = 30 ml
Mt is the final total molarity = 0.0015 molar
Vt is the final volume = V1+V2+V3 = X+20+30 = (X+50) ml

Substitute with the above givens in the equation to get X as follows:
0.003X + 0.0001(20) + 0.002(30) = (X+50)(0.0015)
0.003X + 0.062 = 0.0015X + 0.075
0.003X - 0.0015X = 0.075 - 0.062
0.0015X = 0.013
X = 0.013 / 0.0015 
X = 8.667 ml

The answer is nitrogen-containing bases