Answer is: sodium (Na) and iodine (I₂).
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First ionic bonds in this salt are separeted
because of heat:
</span>NaI(l) → Na⁺(l) + I⁻(l).
Reaction of reduction
at cathode(-): Na⁺(l) + e⁻ → Na(l) /×2.
2Na⁺(l) + 2e⁻ → 2Na(l).
Reaction of oxidation
at anode(+): 2I⁻(l) → I₂(l) + 2e⁻.
The anode is positive
and the cathode is negative.
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Given:
Ammonia
Hydrochloric acid
To find:
The amount excess of reactant left over in ammonia.
Solution:
Ammonia reacts rapidly with hydrochloric acid to form Ammonium Chloride
Equation for the above statement is derived as:
One gram per mole of ammonia
Similarly for 6.91g of
One gram per mole of hydrochloric acid
Similarly for 4.61g of
From the above information we can say that h c l is a limiting reactant.
Limiting reactant is an element that consumes lesser product in a chemical reaction.
Thus the amount of excess reactant is calculated by using the following formula
Amount of excess reactant left over in
Result:
Amount of excess reactant left over ammonia