Based on your question where there is a concentration of 1 M, the weak acid HN02 is 2% ionized nad the PH of the solution is 1.2. The possibilities that could happen to KNO2(s) if it will dissolved into the solution is that w<span>hen more nitrite is added to the solution, the reaction will shift towards the reactants because of La Châtelier's principle. In doing this, the solution will decrease in hydronium concentration and therefore the pH will increase by some amount.</span>
Dissociation of para-bromobenzoic acid can be represented as:
4-BrC₆H₄COOH----->4-BrC₆H₄COO⁻ + H⁺
Dissociation constant of this acid can be calculated as:
Ka={[BrC₆H₄COO⁻][H⁺]}/[4-BrC₆H₄COOH]
as[4-BrC₆H₄COOH=Concentration of para- bromobenzoic acid=0.35 M
And as 1.69% of this acid dissociates to form [BrC₆H₄COO⁻] and [H⁺], so amount of these ions formed will be:
[BrC₆H₄COO⁻]=[H⁺]=1.69×0.35=0.59
So Now Ka=(0.59×0.59)/0.35
=0.99.
- El diafragma, que se localiza debajo de los PULMONES, es el principal músculo de la respiración. Es un músculo largo en forma de domo que se contrae de manera rítmica y continua y, la mayoría del tiempo, de manera involuntaria. En la inhalación, el diafragma se contrae y se allana y la cavidad torácica se amplía.
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Answer:
Explanation:
Naming of the ionic compounds:-
- The name of the cation is written first and the the name of the anion is written after the name of the cation separated by single space.
- The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.
-
In case of transition metals, the oxidation state are written in roman numerals in bracket in front of positive ions.
Hence, given ionic compound:-
Cobalt(II) phosphate
So, Cobalt will have a positive charge of +2
Phosphate is
So, the formula is :-
Co
2 3
Answer:
<h3>The answer is 2.70 g/mL</h3>
Explanation:
The density of a substance can be found by using the formula
From the question
mass = 4.86 g
volume = final volume of water - initial volume of water
volume = 17.3 - 15.5 = 1.8 mL
So we have
We have the final answer as
<h3>2.70 g/mL</h3>
Hope this helps you