Yes. bromine and sodium iodide can react to form sodium bromine and free iodine
Answer:
Expert Verified
Explanation:
For short duration: ... When excess of carbon dioxide gas is passed through lime water then the white precipitate calcium carbonate formed first dissolves due to the formation of a soluble salt calcium hydrogen carbonate (Ca(HCO3)2, and the Solution becomes clear again
Answer: D
Explanation:
A reducing agent is a species that reduces other compounds, and is thereby oxidized. The whole compound becomes the reducing agent. In other words, of a compound is oxidized, then they are the reducing agent. On the other hand, if the compound is reduced, it is an ozidizing agent.
Since we have established that a reducing agent is the compound being oxidized, we know that A is not our answer. An oxidized compound is losing electrons. Choice A states exactly this.
For B, this is true as we have established this already.
C is also correct. Since a reducing agent loses electrons, it becomes more positive. This makes the oxidation number increase.
D would be our correct answer. It is actually a good oxidizing agent is a metal in a high oxidation state, such as Mn⁷⁺.
Molar mass NaOH = 40.0 g/mol
Volume in liters of solution :
5 mL / 1000 => 0.005 L
number of moles :
4 / 40 => 0.1 moles
M = n / V
M = 0.1 / 0.005
= 20 mol/L or 20 M
hope this helps!
Answer:
D
Explanation:
B is solid and liquid while D is liquid and gas
