Answer:
for better understanding and visual understanding
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.
Put a picture we can’t see the arrow
Answer:
1.076 mol (corrected to 2 d.p.)
Explanation:
Take the atomic mass of Ga be 69.7.
since no. of moles = mass/ molar mass,
no. of moles of Ga used = 100.0 / 69.7
= 1.43472023 mol
From the balanced equation, the mole ratio of Ga:S2 = 4:3, which means every 4 moles of Ga can react completely with 3 moles of S2.
So, let the no. of moles of Sulphur required be y.
4 y = 1.43472023 x 3
y = 1.076 mol (corrected to 2 d.p.)
I think it's a radioactive elements I'm only on 6th grade !