Answer:
Molality, Solvent, Solute, Mole fraction, Molarity.
Explanation:
The expression of concentration that provides the moles of solute per kilograms of solvent is Molality. This in the only expression referred to the solvent.
A solution is made up of 0.15 grams of sodium chloride in 1 liter of water. For this solution, the Solvent is water. When water is present, it is usually considered the solvent.
A solution is made up of 0.15 grams of sodium chloride in 1 liter of water. For this solution, the Solute is sodium chloride. There can be 1 or more solutes in a solution.
If you place 5 moles of sodium chloride and 4 moles of sucrose into 11 moles of water, the Mole fraction of sodium chloride would be 0.25. The mole fraction is equal to the moles of a substance divided by the total number of moles.
A way to express concentration that provides the moles of solute per liter of solution is Molarity.
it exists as atom or ion but not as sodium molecule. So there is no any sodium sodium bonding between two sodium atoms. Sodium combines with other element or group to form compounds of sodium like sodium chloride ,sodium carbonate ,sodium sulphate etc. Sodium has symbol Na and its atomic no. is 11 .
Answer:
Explanation:
To convert from moles to grams, we must find the molar mass.
1. Molar Mass
First, identify the elements in the compound. K₂CO₃ It has potassium, carbon, and oxygen. Find these elements and their masses on the Periodic Table.
- K: 39.098 g/mol
- C: 12.011 g/mol
- O: 15.999 g/mol
Note the subscript of 2 after K and 3 after O. We must multiply oxygen's molar mass by 2, then oxygen's by 3, and add carbon.
- 2(39.098 g/mol) + 3(15.999 g/mol) + 12.011 g/mol= 138.204 g/mol
2. Convert Moles to Grams
Use the molar mass as a fraction.
Multiply by the given number of moles: 6.2
There are <u>856.8648 grams</u> of potassium carbonate in 6.2 moles.
a) Alkali metals
=> group 1
=> Li: 1s2 2s => 1s
Na: [Ne] 3s => 3s
K: [Ar] 4s => 4s
Rb: [Kr] 5s => 5s
Cs: [Xe] 6s => 6s
Fr: [Rn] 7s => 7s
=> outer electron configuration is ns, where n is the main energy level: 1, 2, 3, 4, 5, 6,7.
b) Alkaline earth metals
=> group 2 => you have to add 1 electron to the alkaly metal of the same row.
=> Be: [He] 2s2 => 2s2
Mg: [Ne] 3s2 => 3s2
Ca: [Ar] 4s2 => 4s2
Sr: [Kr] 5s2 => 5s2
Ba: [Xe] 6s2 => 6s2
Ra: [Rn[ 7s2 => 7s2
=>outer electron configuration is n s2, where n is the main energy level: 1, 2, 3, 4, 5, 6, 7
c) halogens
=> group 7
=> F: [He] 2s2 2p5 => 2s2 2p5
Cl: [Ne] 3s2 3p5 => 3s2 3p5
Br: [Ar] 3d10 4s2 4p5 => 4s2 4p5
I: [Kr] 4d10 5s2 5p6 => 5s2 5p5
At: [Xe] 4f14 5d10 6s2 6p5
=> outer electron configuration is ns2 np5, where n is the main energy level 1, 2, 3, 4, 5, 6, 7
d) Noble gases
=> group 8
I will show only the outer shell which is what is requested
=> He: 1s2
Ne: ... 2s2 2p6
Ar: ... 3s2 3p6
Kr: ... 4s2 4p6
Xe: ... 5s2 5p6
Rn: ... 6s2 6p6
=> the outer electron configuration is ns2 np6, except for He for which it is 1s2
