Answer:
16% v/v es la nueva concentración de alcohol en la solución
Explanation:
El porcentaje volumen/volumen (% v/v) es definido como 100 veces la relación entre el volumen de soluto (Alcohol en este caso) y el volumen total de la solución (Agua + Alcohol). Para resolver esta pregunta necesitamos hallar el volumen de alcohol y el de agua:
<em>Volumen alcohol:</em>
320cc * (20cc etanol / 100cc) = 64cc etanol
<em>Volumen agua:</em>
80cc + (320cc-64cc) = 336cc agua
<em>% v/v:</em>
64cc / (336cc + 64cc) * 100
= 16% v/v es la nueva concentración de alcohol en la solución
<u>HI :)</u>
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<em>Hope</em><em> this</em><em> helps</em><em> </em><em>:</em><em>)</em>
Greater the Ka value greater is the acid strength. Among these three acids HClO3 is the strongest acid, with greatest Ka and lowest pKa value (-1), Then comes HBrO3, its the second most strongest acid among the three, its Pka value is 0.7, higher than HClO3 but smaller than HIO3 (i.e. 0.77) which the weakest acid among the three.
Answer:
So first thing to do in these types of problems is write out your chemical reaction and balance it:
Mg + O2 --> MgO
Then you need to start thinking about moles of Magnesium for moles of Magnesium Oxide. Based on the above equation 1 mole of Magnesium is needed to make one mole of Magnesium Oxide.
To get moles of magnesium you need to take the grams you started with (.418) and convert to moles by dividing by molecular weight of Mg (24.305), this gives you .0172 moles of Mg.
The theoretical yield would be the assumption that 100% of the magnesium will be converted into Magnesium Oxide, so you would get, based on the first equation, .0172 mol of MgO. Multiplying this by the molecular weight of MgO (24.305+16) gives us .693 g of MgO.
The percent yield is what you actually got in the experiment, and for this you subtract off the total mass from the crucible mass, or 27.374 - 26.687, which gives .66 g of MgO obtained.
Percent yield is acutal/theoretical, .66/.693, or 95.24%.
I'll let you do the same for the second trial, and average percent yield is just an average of the two trials percent yield.
Hope this helps.
