540 nm is the answer ! can you help me out please ? it is a science question add me and message me :)
For NaBr(aq) hydrogen gas is formed at the cathode. Hydrogen in water is reduced to hydrogen gas. For NaBr(aq), bromine, Br2(l), will be produced instead of oxygen gas at the anode.
<span>For sodium sulfate hydrogen gas is formed at the cathode. Hydrogen in water is reduced to hydrogen gas. Oxygen gas will be produced at the anode. </span>
<span>Should someone suggest that sodium metal is formed at the cathode, rest assured that that can't happen. Sodium metal reacts with water to make Na+. </span>
<span>Lead(II) iodide is insoluble in water. Therefore, not much will happen since there is no electrolyte.</span>
Answer:
.125 M
Explanation:
.15 M/L * .125 L = .01875 moles
now dilute to 150 cc (by adding 25 cc)
.01875M / (150/1000) = .125M
Answer:
To move? I think that's right
