<span>Answer:
The HCl and KOH will react until one or the other is gone. As you have a larger volume of an equal concentration of HCl, the KOH will go first.
moles HCl = 0.04000 L * 0.100 M = 0.00400 moles
moles KOH = 0.02500 L * 0.100 M = 0.00250 moles
moles HCl left = 0.00400 - 0.00250 = 0.00150 moles
Your total volume is now 65.00 mL, so the [HCl] = 0.00150 moles / 0.06500 L = 0.0231 M = [H+]
pH = -log [H+] = -log (0.0231) = 1.64</span>
Im pretty sure it would be d.
Answer:
The Bayer process is the principal industrial means of refining bauxite to produce alumina (aluminium oxide) and was developed by Carl Josef Bayer. Bauxite, the most important ore of aluminium, contains only 30–60% aluminium oxide (Al2O3), the rest being a mixture of silica, various iron oxides, and titanium dioxide.
<h3><u>Answer;</u></h3>
0.5 M HBr, pOH = 13.5 ; Has the lowest pH
<h3><u>Explanation;</u></h3>
From the question;
pH = -Log [OH]
or pH = 14 - pOH
Therefore;
For 0.5 M HBr
[H+] = 0.5 M
pH = - Log [0.5]
= 0.30
For; pOH = 13.5
pH = 14 - pOH
= 14 -13.5
= 0.5
For; 0.05 M HCl
pH = - log [H+]
[H+] = 0.05
pH = - Log [0.05]
= 1.30
For; pOH = 12.7
pH = 14 -pOH
= 14 -12.7
= 1.30
For; 0.005 M KOH,
pOH = - log [OH]
[OH-] = 0.005
pOH = - Log 0.005
= 2.30
pH = 14 - 2.30
= 11.7
For; pOH = 2.3
pH = 14 -pOH
= 14- 2.3
= 11.7
Answer is: the approximate freezing point of a 0.10 m NaCl solution is -2x°C.
V<span>an't
Hoff factor (i) for NaCl solution is approximately 2.
</span>Van't Hoff factor (i) for glucose solution is 1.<span>
Change in freezing point from pure solvent to
solution: ΔT = i · Kf · m.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
m - molality, moles of solute per
kilogram of solvent.
</span>Kf and molality for this two solutions are the same, but Van't Hoff factor for sodium chloride is twice bigger, so freezing point is twice bigger.
