Answer:
The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.
Explanation:
Water potential = Pressure potential + solute potential
We have :
C = 0.15 M, T = 273.15 K
i = 1
The water potential of a solution of 0.15 m sucrose= 
(At standard temperature)
The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.
The heat required to raise the temperature of a certain mass of sample to a specific temperature change, we use the formula mCpΔT where m is mass, Cp is the specific heat of the substance and ΔT is the temperature change. In this case, we substitute and form 1.25 g x 0.057 cal/g C *20 C equal to 1.425 calories.
Can you include the pome pls
D. 1 proton and 1 neutron
