A uniform meter stick pivoted at its center has a 155-g mass suspended at the 18.5-cm position and a 25.0-g mass suspended at the 38.4-cm position. At what position should a 275-g mass be suspended
to put the system into equilibrium?

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg

part 1: which substance has a higher solubility?

a

smallest amount left undissolved atall temps in table shown most has dissolved high sol'by