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>
The number of subshells within a certain shell can be identified using the orbital angular momentum quantum number "l"
"l" is given values from zero till (n-1)
So, for n=6
l is given the following values: 0,1,2,3,4,5
Counting the number of subshells, we will find that the shell with n=6 has 6 subshells
Na2SO4(aq)+Ba(NO3)2(aq)------------2NaNO3(aq)+BaSO4(s)
Answer:
8.62 g of Hydrogen
Solution:
Molar mass of CH₄O (Methanol) is 32 g.mol⁻¹.
It means,
32 g of CH₄O contains = 4 g of Hydrogen
Then,
69 g of CH₄O will contain = X g of Hydrogen
Solving for X,
X = (69 g × 4 g) ÷ 32 g
X = 8.62 g of Hydrogen
Answer:
We can do an experiment with vinegar to see what happens to the calcium in an eggshell when it is exposed to an acid.
