Answer: 2Liters
Explanation:
The expression used will be :
where,
= concentration of first antifreeze= 60%
= concentration of second antifreeze= 10%
= volume of first antifreeze = x L
= volume of second antifreeze = 8 L
= concentration of final antifreeze solution= 20%
= volume of final antifreeze = (x+8) L
Now put all the given values in the above law, we get the volume of antifreeze added
Therefore, the volume of 60% antifreeze solution that must be added is 2L
<span>KCl<span>O3</span><span>(s)</span>+Δ→KCl<span>(s)</span>+<span>32</span><span>O2</span><span>(g)</span></span>
Approx. <span>3L</span> of dioxygen gas will be evolved.
Explanation:
We assume that the reaction as written proceeds quantitatively.
Moles of <span>KCl<span>O3</span><span>(s)</span></span> = <span><span>10.0⋅g</span><span>122.55⋅g⋅mo<span>l<span>−1</span></span></span></span> = <span>0.0816⋅mol</span>
And thus <span><span>32</span>×0.0816⋅mol</span> dioxygen are produced, i.e. <span>0.122⋅mol</span>.
At STP, an Ideal Gas occupies a volume of <span>22.4⋅L⋅mo<span>l<span>−1</span></span></span>.
And thus, volume of gas produced = <span>22.4⋅L⋅mo<span>l<span>−1</span></span>×0.0816⋅mol≅3L</span>
Note that this reaction would not work well without catalysis, typically <span>Mn<span>O2</span></span>.
