PH + poH = 14
6.2 +poH = 14
poH = 7.8
Answer:
-179.06 kJ
Explanation:
Let's consider the following balanced reaction.
HCl(g) + NaOH(s) ⟶ NaCl(s) + H₂O(l)
We can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.
ΔH°r = 1 mol × ΔH°f(NaCl(s)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(NaOH(s))
ΔH°r = 1 mol × (-411.15 kJ/mol) + 1 mol × (-285.83 kJ/mol) - 1 mol × (-92.31 kJ/mol) - 1 mol × (-425.61 kJ/mol)
ΔH°r = -179.06 kJ
. How many joules of heat are absorbed to raise
the temperature of 37 grams of water from 25°C
to 75°C?
Answer:
So yes! A mixture can contain just elements if those elements are not chemically reactive to one another. Air is a great example of an elemental mixture as it contains nitrogen, oxygen, and argon (along with some other compounds).
Explanation:
hope this helps you :)
Answer:
pH of resulting solution = 7.98
Explanation:
The balanced equation
HA + NaOH - Na+ + A- + H2O
Number of moles of A = Number of moles of HA = Number of moles of NaOH
= 35.8/1000 * 0.020 = 0.000716 mol
Initial concentration of A = 0.000716/0.0608 = 0.01178 M
pKb = 14 – pKa = 14 -3.9 = 10.1
Kb = 10^{-Kb} = 10^{-10.1} = 7.943 * 10^-11
Kb = [HA][OH-]/[A-]
Kb = a^2/(0.01178 -a) = 7.943 * 10^-11
a^2 + 7.943 * 10^-11 a – 9.357 * 10^-13 = 0
a = 9.673 * 10^-7
OH- = a = 9.673 * 10^-7 M
pOH = -log [OH-] = -log (9.673 * 10^-7) = 6.02
pH = 14-6.02 = 7.98
