3.2 g KClO3
Explanation:
1.1 g C12H22O11 × (1 mol C12H22O11/342.3 g C12H22O11)
= 0.0032 mol C12H22O11
0.0032 mol C12H22O11 × (8 mol KClO3/1 mol C12H22O11)
= 0.026 mol KClO3
Therefore, the minimum amount of KClO3 needed is
0.026 mol KClO3 × (122.55 g KClO3/1 mol KClO3)
= 3.2 g KClO3
<span>The Ksp
of Calcium Hydroxide by Titration of HCl with saturated Ca(OH)<span>2</span></span>
The subscripts tell you <em>how many atoms</em> of an element are in one formula unit of a compound.
If there is no subscript, there is only one atom of the element.
In one formula unit of Na₂PO₃F, there are
Two atoms of sodium (Na)
One atom of phosphorus (P)
Three atoms of oxygen (O)
One atom of fluorine (F)
Answer:
See explanation below
Explanation:
In this case, HCl is a strong acid, therefore, it dissociates completely in solution.
To know the quantity of water we need to add, we first need to know the concentration of the acid with pH = 6:
[H⁺] = antlog(-pH)
[H⁺] = antlog(-6) = 1x10⁻⁶ M
This means that the concentration is being diluted.
Now, even if we add great quantities of water, and the concentration and volume change, there is one time that do not change despite the quantity of water added; this is the moles. So, all we have to do, is calculate the moles of the acid in the 1 mL of water, and then, the volume of the acid when it's dilluted:
moles HCl = 0.1 * (1/1000) = 1x10⁻⁴ moles
Now that we have the moles, we can calculate the volume which the acid with the lowest concentration has:
V = mol/M
V = 1x10⁻⁴ / 1x10⁻⁶
V = 100 L
This means that we need to add 99.999 mL of water
