The mass of water is the same as the mass of ice
Answer:
Lose two electrons.
Explanation:
Barium is present in group 2.
It is alkaline earth metal.
Its atomic number is 56.
Its electronic configuration is Ba₅₆ = [Xe] 6s².
In order to attain the noble gas electronic configuration it must loses its two valance electrons.
When barium loses it two electron its electronic configuration will equal to the Xenon.
The atomic number of xenon is 54 so barium must loses two electrons to becomes equal to the xenon.
Nitrogen fixation means form nitrits and dentrification means nitrits will not formed
Answer:
c Example 1 represents a liquid and Example 2 represents a solid.
Explanation:
Example 1: Young children dancing slowly around one another
The young children dancing slowly around one another can be pictured as the flow of liquid. In liquids, the molecules are held about weakly and they slide on top of each other. They are held by weak attractive forces. This is clear picture of a liquid.
Example 2: Newborn babies sitting in their given spots in a crib
This is clear example of a solid. In a solid the molecules are held about a fixed spot. The attractive forces in liquids is very great and the molecules therein do not move about randomly.
Answer:
Option d.
1 mole AlCl3in 500 g water
Explanation:
ΔT = Kf . m . i
Freezing T° of solution = - (Kf . m . i)
In order to have the lowest freezing T° of solution, we need to know which solution has the highest value for the product (Kf . m . i)
Kf is a constant, so stays the same and m stays also the same because we have the same moles, in the same amount of solvent. In conclussion, same molality to all.
i defines everything. The i refers to the Van't Hoff factor which are the number of ions dissolved in solution. We assume 100 & of ionization so:
a. Glucose → i = 1
Glucose is non electrolytic, no ions formed
b. MgF₂ → Mg²⁺ + 2F⁻
i = 3. 1 mol of magnessium cation and 2 fluorides.
c. KBr → K⁺ + Br⁻
i = 2. 1 mol potassium cation and 1 mol of bromide anion
d. AlCl₃ → Al³⁺ + 3Cl⁻
i = 4. 1 mol of aluminum cation and 3 mol of chlorides.
Kf . m . 4 → option d will has the highest product, therefore will be the lowest freezing point.
