Using the solubility curve above, what is the solubility of KBr when the temperature is 80°C?
A)
86 g/100 g water
90 g/100 g water
96 g/100 g water
D)
102 g/100 g water

96 g/100 g water

Explanation:

Kindly find the graph attached.

A solubility curve expresses how soluble a solute is at a given temperature in a particular solvent.

Solubility varies with temperature changes.

The x-axis denotes temperature and the y-axis is the solubility axis.

The solubility curve of KBr is the green line. Locate the given temperature on the x-axis and find the y-ordinate when it hits the curve.

The value gives solubility at that temperature.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft