Answer:
Option D. 1×10¯¹⁰ M
Explanation:
From the question given above, the following data were obtained:
pH = 10
Hydrogen ion concentration, [H+] =?
The hydrogen ion concentration, [H+] of the Ba(OH)2 solution can be obtained as follow:
pH = – Log [H+]
pH = 10
10 = – Log [H+]
Divide both side by – 1
– 10 = Log [H+]
Take the antilog of – 10
[H+] = antilog (– 10)
[H+] = 1×10¯¹⁰ M
Therefore, the hydrogen ion concentration, [H+] of the Ba(OH)2 solution is 1×10¯¹⁰ M
The answer is most likely that the answer is B. Because Space is a dangerous place in which you need certain things to explore it. If you want to go to to Jupiter, for example you need lots of Oxygen, fuel, a good rocket, willing people and many more things, but there is a way technology can skip all that. the technology we have now is not capable of doing that though.
Answer:
See Explanation
Explanation:
Let us consider the ionization of these compounds;
H3PO4 ⇔3H^+ + PO4^3-
H3BO3 ⇔3H^+ + BO3^3-
The next to consider is the type of electrolyte the both solution are; the both solutions are weak electrolytes and weak electrolytes do not ionize to a large extent.
The implication of this is that, not so much number of ions is added to the solution due to the poor ionization of these weak electrolytes. Hence, in spite of the subscript of 3, the conductivity of the solution does not significantly improve for the reason stated here quite unlike when strong electrolytes are used.
We are given that the specific heat of water is 4.18 J / g
°C. We know that the molar mass of water is 18.02 g/mol, therefore the molar
heat capacity is:
molar heat capacity = (4.18 J / g °C) * 18.02 g / mol
<span>molar heat capacity = 75.32 J / mol °C</span>
Answer:
Explanation:
glucose-1-phosphate → glucose-6-phosphate, ΔGo = -7.28 kJ/mol
fructose-6-phosphate → glucose-6-phosphate, ΔGo = -1.67 kJ/mol
subtracting the equation
glucose-1-phosphate - fructose-6-phosphate = 0 , ΔGo = -7.28 - ( -1.67 ) kJ / mol
glucose-1-phosphate = fructose-6-phosphate ΔGo = - 5.61 kJ / mol
- ΔGo = RT lnK
5.61 x 10³ = 8.31 x 298 x lnK
lnK = 2.265
K = 9.63 .