The pH of an aqueous solution that has a concentration of 0.35 M NaF and pKa for HF = 3.14 is 3.6.
<h3>How to calculate pH?</h3>
The pH of a solution refers to the degree of acidity or alkalinity of the solution. It can be calculated using the Henderson-Hasselbalch Equation as follows:
pH = pka + log ([A-]/[HA])
Where;
- A- = conjugate base
- HA = weak acid
pH = pKa + log([F-]/[HF])
pH = 3.14 + log(1/0.35)
pH = 3.14 + 0.4559 = 3.595
Therefore, the pH of an aqueous solution that has a concentration of 0.35 M NaF and pKa for HF = 3.14 is 3.6.
Learn more about pH at: brainly.com/question/15289741
Sodium has has 11 protons, 11 electrons and 12 neutrons.
<h3>What is the periodic table?</h3>
The periodic table shows the elements as they are arranged in increasing atomic number. The elements on the left hand side of the table are metals. Those on the right hand side of the table are mostly nonmetals.
A element in the first column is sodium;
The element Na is called sodium and it belongs to group 1 elements. It has 11 protons, 11 electrons and 12 neutrons.
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Answer:
ΔH = 125.94kJ
Explanation:
It is possible to make algebraic sum of reactions to obtain ΔH of reactions (Hess's law). In the problem:
1. 2W(s) + 3O2(g) → 2WO3(s) ΔH = -1685.4 kJ
2. 2H2(g) + O2(g) → 2H2O(g) ΔH = -477.84 kJ
-1/2 (1):
WO3(s) → W(s) + 3/2O2(g) ΔH = 842.7kJ
3/2 (2):
3H2(g) + 3/2O2(g) → 3H2O(g) ΔH = -716.76kJ
The sum of last both reactions:
WO3(s) + 3H2(g) → W(s) + 3H2O(g)
ΔH = 842.7kJ -716.76kJ
<h3>ΔH = 125.94kJ </h3>
Answer:
it is nanometers the third one
Explanation:
hope this helps
