2.79 mol of MnSiO3
This will be very close but may not be exactly accurate, depending on what your periodic table says.
Answer:
c
Explanation:
iodine is the answer.please mark me as brianlist.
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.
Question:
What are advantages of using Uranium as an energy source?
Answer(s):
-Small amounts of Uranium generate large amounts of energy
-Uranium occurs in huge reserves
-It has a longer lifetime than other non-renewable sources of energy
-Brainly Answerer
Answer:
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