The given question is incomplete. The complete question is as follows.
If more sugar were added to the solution on the left side of the tube, what would happen to the water level on the right side of the tube?
- There is more solute in the left arm of the tube than in the right arm of the tube.
- There is less free water in the right arm of the tube than in the left arm of the tube.
- Water is tightly clustered around the hydrophilic solute molecules on both sides of the membrane.
- Solutes can pass through the selectively permeable membrane from right to left, but free water cannot.
Explanation:
It is known that when water is able to diffuse through a semi-permeable membrane from a region of high water concentration to low water concentration is called osmosis.
Since, more amount of solute is present on the right side of the tube so, there will be less amount of free water present over there as compared to the left side of the tube.
Also, water will be clustered around the hydrophilic solute molecules which is present on both sides of the membrane.
Thus, we can conclude that the true statements are as follows.
- There is less free water in the right arm of the tube than in the left arm of the tube.
- Water is tightly clustered around the hydrophilic solute molecules on both sides of the membrane.
The more shielding, inner electrons
Answer:
potential energy (im positive that its right)
Answer:
0.437
Explanation:
Divide the length value by 1000 to get the answer.
Answer:
pH = 12.2
Explanation:
Given data:
Mass of lime = 0.69 g
Volume = 1535 mL (1535 / 1000 = 1.535 L)
pH of solution = ?
Solution:
First of all we will determine the molarity.
Molarity = moles of solute / volume in litter
Number of moles = mass / molar mass
Number of moles = 0.69 g / 56.1 g/mol
Number of moles = 0.0123 mol
Molarity = moles of solute / volume in litter
Molarity = 0.0123 mol / 1.535 L
Molarity = 0.008 M
One mole of CaO neutralize two mole of OH⁻.
0.008 M×2 = 0.016 M
[OH⁻] = 0.016 M
pOH = -log [OH⁻]
pOH = [0.016]
pOH = 1.8
14 = pH +pOH
pH = 14 - pOH
pH = 14-1.8
pH = 12.2