Explanation:
The only flaw I can find is you squared 3 instead of cubing it and it will be 27X^4 instead of 9x^4.
This reduces the amount slightly, but the number is still incredibly high (about 10 ^ 5 L is what I've calculated). Your professor might want to point out that this will not be a effective experiment due to the large volume of saturated
The Ksp value of Ca(OH)2 on the site (I used 5.5E-6 [a far more soluble compound than Al(OH)3]) and estimated how much of it will be needed. My calculation was approximately 30 ml. If you were using that much in the experiment, it implies so our estimates for Al(OH)3 are right, that the high amount is unreasonably big and that Al(OH)3 will not be a suitable replacement unless the procedure was modified slightly.
The part of the experiment that’s is not touched by the independent variable and is for comparison is called the :
Control Group
E is Bohrs model the dots on the rings represent the valence electrons
Answer:
Mole fraction of 
= 0.58
Mole fraction of 
= 0.42
Explanation:
Let the mass of and = x g
Molar mass of = 33.035 g/mol
The formula for the calculation of moles is shown below:
Thus,
Molar mass of = 46.07 g/mol
Thus,
So, according to definition of mole fraction:
Mole fraction of = 1 - 0.58 = 0.42
Answer:
the answer is 10 times
Explanation:
because it takes 10 times as much energy -3330 j - to melt 10.0 grams of ice.
