Answer:
D. +5.7 kJ/mol
Explanation:
Molar free energy (ΔG) in the transportation of uncharged molecules as glucse through a cell membrane from the exterior to the interior of the cell is defined as:
ΔG = RT ln C in / C out
knowing R is 8,314472 kJ/molK; T is 298K Cin = 200mM and Cout = 20mM
ΔG = 5,7 kJ/mol
Right answer is:
D. +5.7 kJ/mol
I hope it helps!
The change is that the water will freeze to 0 or minus I don’t know as I’m not to sure
Answer:
the pressure would be 0.9 atmospheres
Explanation:
you just gotta add the presures for each of the gases that are added
Answer:
Base on the properties the substances possess the substance change phase from gas to liquid.
Explanation:
Generally, matter can exist in three phase namely liquid , solid and gas. The scientist has a container with a substance inside.
At first the substance moves away from each other . This means the substance was first in the gas phase . Gas molecules, because of the energy they possess , it can move rapidly and randomly and most at times move away from each other . The gas molecules tends to fill the whole volume of the container and the shape of gases are indefinite.
Later the molecules move around each other. This is a property of a liquid . A substance in liquid phase have the ability to move freely but they stay together because of the force of attraction holding them together . The substance only have the ability to move around each other because the forces holding them together won't allow them to move apart.
Base on the properties the substances possess the substance change phase from gas to liquid.
Answer:
2.7 °C.kg/mol
Explanation:
Step 1: Calculate the freezing point depression (ΔT)
The normal freezing point of a certain liquid X is-7.30°C and the solution freezes at -9.9°C instead. The freezing point depression is:
ΔT = -7.30 °C - (-9.9 °C) = 2.6 °C
Step 2: Calculate the molality of the solution (b)
We will use the following expression.
b = mass of solute / molar mass of solute × kilograms of solvent
b = 102. g / (162.2 g/mol) × 0.650 kg = 0.967 mol/kg
Step 3: Calculate the molal freezing point depression constant Kf of X
Freezing point depression is a colligative property. It can be calculated using the following expression.
ΔT = Kf × b
Kf = ΔT / b
Kf = 2.6 °C / (0.967 mol/kg) = 2.7 °C.kg/mol
