The labels the table to indicate when each statement Is true. Labels can be used once, more than once, or not at all, The orange dye moves independently of the purple dye. 2. Concentration gradients exist that drive diffusion of both dyes. : <u>free water, solute, free water, solute.</u>
In chemistry, attention is the abundance of a constituent divided by way of the total quantity of an aggregate. several types of mathematical descriptions may be prominent: mass concentration, molar concentration, range concentration, and quantity concentration.
it's miles the amount of solute dissolves in one hundred g solvent. If the attention of the answer is 20 %, we understand that there are 20 g solutes in one hundred g solution. instance: 10 g salt and 70 g water are mixed and the solution is ready. find awareness of the answer by means of percentage mass.
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Answer:
D.) the density changed
Explanation:
This would be the answer because it is a specific amount of Nitrogen undergoing a physical change. Since the color and boiling point don't influence a change in volume, those are ruled out. Also, nitrogen's molecular mass never changes, so that would not make sense either. But if the density, which is mass/volume, has changed (more specifically decreases) then it means the volume has increased.
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From the information in the question, the E° and E for the cell is 0.00 V and 0.12 V.
Using the Nernst equation;
Ecell = E°cell - 0.0592/n log Q
We know that E°cell = 0.00 V since the anode and cathode are both made up of cadmium.
Substituting the given values into the Nernst equation;
Ecell = 0.00 V - 0.0592/2 log (1.0 × 10-5 M/0.100 M)
Ecell = 0.00 V - 0.0296 log(1 × 10^-4)
Ecell = 0.12 V
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Answer: Option (a) is the correct answer.
Explanation:
The given data is as follows.
= 4.19 
= 1.9
Heat of vaporization (
) at 1 atm and 
is 2259 kJ/kg
= 0
Therefore, calculate the enthalpy of water vapor at 1 atm and as follows.
= +
= 0 + 2259 kJ/kg
= 2259 kJ/kg
As the desired temperature is given 
and effect of pressure is not considered. Hence, enthalpy of liquid water at 10 bar and is calculated as follows.
= 
= 334.781 kJ/kg
Hence, enthalpy of water vapor at 10 bar and is calculated as follows.
= 
= 2410.81 kJ/kg
Therefore, calculate the latent heat of vaporization at 10 bar and as follows.
= 
= 2410.81 kJ/kg - 334.781 kJ/kg
= 2076.029 kJ/kg
or, = 2076 kJ/kg
Thus, we can conclude that at 10 bar and latent heat of vaporization is 2076 kJ/kg.
Answer : A
Explanation:
Answer A is correct I just did the assignment.
