Answer:
Back titration is a titration done in reverse; instead of titrating the original sample, a known excess of standard reagent is added to the solution, and the excess is titrated.
I hope it's helpful!
Answer:
69.4kg
Explanation:
mole =volume/1000 * mole conc.
=1250/1000 * 0.75
mole =0.9
:• mole =mass /molar mass
mass =0.9*[40+32+2]
=69.4kg
Answer:
Group 1: LiOH, NaOH, KOH
Group 2: Ca(OH)2, Sr(OH)2, Ba(OH)2
Explanation:
Hey there!:
Mole ratio :
2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O
2 moles C₂H₆ -------------------- 7 moles O₂
3.0 moles C₂H₆ ------------------ moles O₂ ??
moles O₂ = 3.0 x 7 / 2
moles O₂ = 21 / 2
moles O₂ = 10.5 moles of O₂
Hope this helps!
Answer:
<em>The pH of the solution is 7.8</em>
Explanation:
The concentration of the solution is 0.001M and the dye could be in its protonated and deprotonated forms. If the concentration of the protonated form [HA] is 0.0002 M the concentration of the deprotonated form will be the subtraction between the concentration of the bye and the concentration of the protonated form:
[A-] = 0.001M - 0.0002M = 0.0008M
Also, the Henderson-Hasselbalch equation is
this equation shows the dependency between the pH of the solution, the pKa and the concentration of the protonated and deprotonated forms. Thus, replacing in the equation