I believe the answer for this is:
C. Visible Light Rays
I hope this helps! :D
Answer:
adjusting the air mix by rotating the barrel and adjusting the gas with the needle valve to obtain a flame of suitable height and intensity
Br2 because it is non polar and so is CCl4 and like molecules dissolve like molecules therefore Br2 will dissolve in CCl4
Answer:
1.427x10^-3mol per L
Explanation:

I could use ⇌ in the math editor so I used ----
from the question each mole of Y(IO3)3 is dissolved and this is giving us a mole of Y3+ and a mole of IO3^3-
Ksp = [Y^3+][IO3-]^3
So that,
1.12x10^-10 = [S][3S]^3
such that
1.12x10^-10 = 27S^4
the value of s is 0.001427mol per L
= 1.427x10^-3mol per L
so in conclusion
the molar solubility is therefore 1.427x10^-3mol per L