Answer:
electron transfer to occur to form ions.
Answer:
- <u>Yes, it is 14. g of compound X in 100 ml of solution.</u>
Explanation:
The relevant fact here is:
- the whole amount of solute disolved at 21°C is the same amount of precipitate after washing and drying the remaining liquid solution: the amount of solute before cooling the solution to 21°C is not needed, since it is soluble at 37°C but not soluble at 21°C.
That means that the precipitate that was thrown away, before evaporating the remaining liquid solution under vacuum, does not count; you must only use the amount of solute that was dissolved after cooling the solution to 21°C.
Then, the amount of solute dissolved in the 600 ml solution at 21°C is the weighed precipitate: 0.084 kg = 84 g.
With that, the solubility can be calculated from the followiing proportion:
- 84. g solute / 600 ml solution = y / 100 ml solution
⇒ y = 84. g solute × 100 ml solution / 600 ml solution = 14. g.
The correct number of significant figures is 2, since the mass 0.084 kg contains two significant figures.
<u>The answer is 14. g of solute per 100 ml of solution.</u>
Answer:
1.75M
Explanation:
The problem deals with finding the molarity of the given compound.
The compound is:
Na₂SO₄.10H₂O
Mass of the compound = 10.5kg = 10500g
Volume of the compound = 18.6L
Molarity is the number of moles of solute in a solution;
Molarity = 
Number of moles = 
Molar mass of Na₂SO₄.10H₂O = 2(23) + 32 + 4(16) + 10[2(1) + 16]
= 322g/mol
Now;
Number of moles = 
= 32.6mole
So;
Molarity = 
= 1.75M
Answer:
a) Carbon.
Explanation:
The organic chemistry is normally based on the chemical elements Carbon, Hydrogen,oxygen, sulphur and Nitrogen. Sometimes a mixture or compound of one or more of these will be found such as (carbohydrates, fats, proteins). The varieties of organic molecules are based on these atoms, with carbon as the skeleton of the organic molecule.
Hence out of the four options given we find that though nitrogen, sulphur are also common in organic the most common is the first option
a) Carbon.
