The pH of a solution is 9.02.
c(HCN) = 1.25 M; concentration of the cyanide acid
n(NaCN) = 1.37 mol; amount of the salt
V = 1.699 l; volume of the solution
c(NaCN) = 1.37 mol ÷ 1.699 l
c(NaCN) = 0.806 M; concentration of the salt
Ka = 6.2 × 10⁻¹⁰; acid constant
pKa = -logKa
pKa = - log (6.2 × 10⁻¹⁰)
pKa = 9.21
Henderson–Hasselbalch equation for the buffer solution:
pH = pKa + log(cs/ck)
pH = pKa + log(cs/ck)
pH = 9.21 + log (0.806M/1.25M)
pH = 9.21 - 0.19
pH = 9.02; potential of hydrogen
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Molar mass of
<u>As we know</u>–
1 mol = formula units
1 mol = 143.092 g = formula units
Henceforth –
formula units –
Glycolysis yields 2 ATP molecules, Krebs cycle yields 2 ATP molecules, ETS yields 34 ATP molecules.
Answer:
C. fluorine (F) and chlorine (Cl)
D. arsenic (As) and antimony (Sb)
Explanation:
In the periodic table , all the elements are arranged according to the atomic number ,
and the elements are placed in groups and periods ,
The elements with similar chemical and physical properties are placed in a common group .
The elements present in the same group have the same number of valence electrons in the valence shell .
Hence , from the given options ,
fluorine (F) and chlorine (Cl) belongs to group 17 with 7 valence electrons in the outermost shell .
arsenic (As) and antimony (Sb) belong to group 15 with 3 valence electrons in the outermost shell .