<span>B. S⁰(s) + 2H⁺ + 2e⁻ –--> H2S⁰(g)
by mass: 1 S and 2 H ----> 2 H and 1S True.
by charge : 0 +(2*(+1)) + 2*(-1) = 0, 0+2-2=0, 0 = 0 True.</span>
Answer:
um.. is this a quiz or what girl...
Answer:
Explanation:
Initial burette reading = 1.81 mL
final burette reading = 39.7 mL
volume of NaOH used = 39.7 - 1.81 = 37.89 mL .
37.89 mL of .1029 M NaOH is used to neutralise triprotic acid
No of moles contained by 37.89 mL of .1029 M NaOH
= .03789 x .1029 moles
= 3.89 x 10⁻³ moles
Since acid is triprotic , its equivalent weight = molecular weight / 3
No of moles of triprotic acid = 3.89 x 10⁻³ / 3
= 1.30 x 10⁻³ moles .
<span>0.002 moles is the answer </span>