Sulfur is a non-metal, being in the third period, in group 6
That’s a dumb question, because It depends on the experiment. I would guess “a” or “d” because in most cases running out of time during a lab, or getting impatient, etc can give you a lower yield.
Unless I’m misreading “d”, it just seems like a more in-depth version of “a”. So it wouldn’t hurt to try that one.
The answer would be 20% (B)
Answer:
Explanation:
mole of NaOH present = molarity x volume
= 1.0 X 0.05 = 0.05 mole
<em>Recommended mole of HCl </em>= 1.1 x 0.05 = 0.055
<em>Mole of HCl carelessly added by Jacob </em>= 1.1 x 0.04 = 0.044
From the equation of reaction:
HCl + NaOH ----> NaCl + H2O
The ratio of mole of HCl to that of NaOH for a complete neutralization reaction is 1:1. However, the recommended mole of HCl (0.055 mole) is more than the mole of NaOH (0.05 mole). <u>Hence, the recommended endpoint of the reaction is supposed to be acidic.</u>
The mole of HCl added by Jacob (0.044) is short of the recommended amount (0.055) and also short of the amount required for a neutral endpoint (0.05). <u>This means that the endpoint will have an excess amount of NaOH and as such, basic instead of the desired acidic endpoint.</u>
