SrSo4 = Sr(2+) + SO4(2-)
Let’s say that the initial concentration of SrSo4 was 1. ( or we have 1 mole of this reagent).
When The reaction occurs part of SrSo4is dissociated. And we get X mole Sr(2+) and So4(2-).
Ksp=[Sr(2+)]*[SO4(2-)]
X^2=3.2*10^-7
X=5.6*10^-4
Answer:
a.labeling food with an ingredient list to inform consumers
Explanation:
Answer:
It works because of the significant principle of piezoelectricity
Explanation:
Answer:
Yes. Example: <u>Sulfur hexafluoride (SF₆) molecule</u>
Explanation:
According to the octet rule, elements tend to form chemical bonds in order to have <u>8 electrons in their valence shell</u> and gain the stable s²p⁶ electronic configuration.
However, this rule is generally followed by main group elements only.
Exception: <u>SF₆ molecule</u>
In this molecule, six fluorine atoms are attached to the central sulfur atom by single covalent bonds.
<u>Each fluorine atom has 8 electrons in their valence shells</u>. Thus, it <u>follows the octet rule.</u>
Whereas, there are <u>12 electrons around the central sulfur atom</u> in the SF₆ molecule. Therefore, <u>sulfur does not follow the octet rule.</u>
<u>Therefore, the SF₆ molecule is known as a </u><u>hypervalent molecule</u><u> or expanded-valence molecule.</u>
We are given
0.2 M HCHO2 which is formic acid, a weak acid
and
0.15 M NaCHO2 which is a salt which can be formed by reacting HCHO2 and NaOH
The mixture of the two results to a basic buffer solution
To get the pH of a base buffer, we use the formula
pH = 14 - pOH = 14 - (pKa - log [salt]/[base])
We need the pKa of HCO2
From, literature, pKa = 1.77 x 10^-4
Substituting into the equation
pH = 14 - (1.77 x 10^-4 - log 0.15/0.2)
pH = 13.87
So, the pH of the buffer solution is 13.87
A pH of greater than 7 indicates that the solution is basic and a pH close to 14 indicates high alkalinity. This is due to the buffering effect of the salt on the base.
