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Elena-2011 [213]
3 years ago
10

Whats the ph for a solution poh4 9.78 concentration of solution

Chemistry
1 answer:
MrMuchimi3 years ago
3 0

Answer:

4.22

Explanation:

According to the question, the pOH of the solution is 9.78. You may recall that pOH is the hydroxide concentration of a solution.

Also pOH = -log[OH^-]. Hence the pOH is obtained from the hydroxide ion concentration.

Finally, pH + pOH =14

Hence;

pH = 14-pOH

pH= 14-9.78 = 4.22

pH= 4.22

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Usually potassium hydrogen phthalate is kept very pure. But Stu Dent thinks the bottle of potassium hydrogen phthalate has been
Ahat [919]

Answer:

1.784 g

Explanation:

The equation of the reaction is;

NaOH(aq) + KHC8H4O4(aq) --------> KNaC8H4O4(aq) + H2O(l)

Number of moles of NaOH reacted = 17.47/1000 * 0.5000 M

Number of moles of NaOH reacted =8.735 * 10^-3 moles

From the reaction equation;

1 mole of NaOH reacted with 1 mole of KHC8H4O4

Hence, 8.735 * 10^-3 moles of NaOH reacts with 8.735 * 10^-3 moles of KHP.

So,

Mass of KHP reacted = 8.735 * 10^-3 moles * 204.2 g/mol = 1.784 g

5 0
3 years ago
3. You are asked to prepare an aqueous solution. What would be a better choice for a solute potassium nitrate or
liubo4ka [24]

Answer:

i do not knowwww

Explanation:

8 0
3 years ago
What is the main factor that prevents the growth of tropical plants in the northern part of the us
Inessa05 [86]
Extreme weather e.g. snow and cold weather. Another main contributing factor is that many places have a long dry season.
3 0
3 years ago
One gram of liquid benzene is burned in a bomb calorimeter. The temperature before ignition was 20.826 C, and the temperature af
Licemer1 [7]

Answer:

fH = - 3,255.7 kJ/mol

Explanation:

Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.

Qsystem + Qcombustion = 0

Qsystem = heat capacity*ΔT

10000*(25.000 - 20.826) + Qc = 0

Qcombustion = - 41,740 J = - 41.74 kJ

So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:

n = mass/molar mass = 1/ 78

n = 0.01282 mol

fH = -41.74/0.01282

fH = - 3,255.7 kJ/mol

4 0
4 years ago
A certain reaction is thermodynamically favored at temperatures below 400. K, but it is not favored at temperatures above 400. K
Ilya [14]

Answer:

-0.050 kJ/mol.K

Explanation:

  • A certain reaction is thermodynamically favored at temperatures below 400. K, that is, ΔG° < 0 below 400. K
  • The reaction is not favored at temperatures above 400. K, that is. ΔG° > 0 above 400. K

All in all, ΔG° = 0 at 400. K.

We can find ΔS° using the following expression.

ΔG° = ΔH° - T.ΔS°

0 = -20 kJ/mol - 400. K .ΔS°

ΔS° = -0.050 kJ/mol.K

8 0
3 years ago
Read 2 more answers
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