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.
Answer:
1600
Explanation:
5×10^4÷2.5×10^2
(5×10^4)
(10^4)
(5×40)
(200)
(200÷2.5)
(80)
(80×10^2)
(10^2)
(20)
(80×20)
Answer is 1600.
Sorry if it's not correct.
Not sure. How to answer this question.Look it up
Answer:
Frecuency = 5,83x10⁻⁷ Hz
Explanation:
The equation that connects wavelenght and frequency is given by:
λ = c/ν
λ=wavelenght (expressed in lenght´s units)
c= speed of light (3x10⁸ m/sec)
ν=frequency (expressed in units of time⁻¹ or Herzt)
In our case, λ=5,14x10⁻⁷ m , so replacing in our previous formula, this gives us the final result of ν (frequency for green light) of 5,83x10¹⁴ Hz (or Herzt)
