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.
Yes, Bobby is correct
Explanation:
Anomalously high boiling point of water is as a result of the intermolecular forces between the molecules of water.
The intermolecular forces found in water are the very strong hydrogen bonds. The bulk of the physical properties of matter are due to the intermolecular forces that they possess.
- Hydrogen bonds are stronger than van der waals forces and they are more effective in binding molecules together into larger units.
- Substances whose molecules join via hydrogen bonds have higher boiling points i.e lower volatility than those with van der waals forces.
- Hydrogen bond is actually an electrostatic attraction between hydrogen atom of none molecule and the electronegative atom(O or N or F) of a neighboring molecule.
Learn more:
Hydrogen bonds brainly.com/question/10602513
#learnwithBrainly
Hello!
The compound that forms from potassium and chromate is Potassium Chromate (K₂CrO₄)
This compound has a yellow lime color and is used in enamels, leather finishes, stainless steels and for the identification of silver ion in a silver nitrate solution. It is a powerful oxidizing agent and is produced by the treatment of potassium chromate with potassium hydroxide. It is a very toxic compound an is a known carcinogenic agent.
Have a good day!
A solution is a mixture of two or more substances combined so that it is uniform, which means you cannot see the components. <span />
Answer:
Here's what I get.
Explanation:
The MO diagrams of KrBr, XeCl, and XeBr are shown below.
They are similar, except for the numbering of the valence shell orbitals.
Also, I have drawn the s and p orbitals at the same energy levels for both atoms in the compounds. That is obviously not the case.
However, the MO diagrams are approximately correct.
The ground state electron configuration of KrF is
KrF⁺ will have one less electron than KrF.
You remove the antibonding electron from the highest energy orbital, so the bond order increases.
The KrF bond will be stronger.
