Answer:
two may be the answer (2)
Answer:
27.03 %
Explanation:
The formula for the calculation of moles is shown below:
Given: For calcium
Given mass = 10.0 g
Molar mass of calcium = 40.078 g/mol
<u>Moles of calcium = 10.0 g / 40.078 g/mol = 0.2495 moles</u>
According to the given reaction:
1 mole of calcium on reaction forms 1 mole of calcium hydroxide
Thus,
0.2495 moles of calcium on reaction forms 0.2495 mole of calcium hydroxide
<u>Moles of calcium hydroxide = 0.2495 moles</u>
Molar mass of calcium hydroxide = 74.093 g/mol
Thus, Mass = Moles * Molar mass = 0.2495 moles * 74.093 g/mol = 18.5 g
<u>Theoretical yield = 18.5 g</u>
<u>
Given experimental yield = 5.00 g
</u>
<u>% yield = (Experimental yield / Theoretical yield) × 100 = (5.00/18.5) × 100 = 27.03 %
</u>
It is lead (ll) iodide or just lead iodide
We are asked to provide the net ionic equation for the reaction of HF (aq) and NaF (aq). HF is a weak acid and is in the following equilibrium:
HF (aq) ⇄ H⁺ (aq) + F⁻ (aq)
Meanwhile, NaF (aq) is an ionic compound that will dissociate completely in aqueous solutions:
NaF (aq) → Na⁺ (aq) + F⁻ (aq)
We can combine the ionic species with HF, as we are told to show F⁻ as a reactant:
HF (aq) + Na⁺ (aq) + F⁻ (aq) → HF (aq) + Na⁺ (aq) + F⁻ (aq)
We can eliminate the spectator ions, which in this case are Na⁺ ions, and this leaves us with the net ionic equation involving F⁻:
HF (aq) + F⁻ (aq) → HF (aq) + F⁻ (aq)
In this instance, the proton is just transferred between F⁻ ions and the end result is the formation of more HF, so there is no net reaction taking place.