Answer:
pH = 5.47
Explanation:
The equilibrium that takes place is:
HIO ↔ H⁺ + IO⁻
Ka = ![\frac{[H+][IO-]}{[HIO]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH%2B%5D%5BIO-%5D%7D%7B%5BHIO%5D%7D)
= 2.3 * 10⁻¹¹
At equilibrium:
<u>Replacing those values in the equation for Ka and solving for x:</u>
Then [H⁺]=3.39 * 10⁻⁶, thus pH = 5.47
Answer:
B
Explanation:
the group number is=valence electrons. element 1 is in group 1 element 18 is in group 8. 1<8
During the reaction of glucose and fructose with excess phenylhydrazine to form osazone, only the C-1andC-2 atoms of glucose and fructose participate in the reaction. The rest of the molecule remains intact. Hence, glucose and fructose produce the same osazone.
Answer:
the ion present in the original solution is Ca2+
Explanation:
Precipitation reactions occur when cations and anions in aqueous solution combine to form an insoluble ionic solid called a precipitate.
<u>Step1</u> : If we add Nacl to the solution, there is no precipitate formed
⇒The only possible ion that can form a precipate with Cl- is Ag+; since there is no precipitate formed, Ag+ is not present
<u>Step2</u> : If we add Na2SO4 to the solution, a white precipitate is formed
The possible ions to bind at SO42- are Ca2+ and Fe2+
But the white precipitate formed, points in the direction of Ca2+
⇒This means calcium is present
<u>Step3</u> : If we add Na2CO3 to the filtered solution, there is a precipate formed
Ca2+ will bind also with CO32- and form a precipitate
So the ion present in the original solution is Ca2+
