First, we write out a balanced equation.
HA <--> H(+) + A(-)
Next, we create an ICE table
HA <--> H+ + A-
[]i 0.40M 0M 0M
Δ[] -x +x +x
[]f 0.40-x x x
Next, we write out the Ka expression.
Ka = [H+][A-]/[HA]
Ka = x*x/(0.40-x)
However, because Ka is less than 10^-3, we can assume the amount of dissociation is negligible. Thus,
Assume 0.40-x ≈ 0.40
Therefore, 1.2x10^-6 = x^2/0.40
Then we solve for the [H+] concentration, or x
x=6.93x10^-4
Next, to find pH we do
pH = -log[H+]
pH = -log[6.93x10^-4]
pH = 3.2
Answer:
1) B
2) D
3) A
4) Ga
5) K
6)Po
7) Atomic size increases down the group
8) B<Al<Ga<In<Tl
9)Se<C<Ga
10) ionization energy is the energy required to remove electrons from the outermost shell of an atom.
Explanation:
In the periodic table, the properties of elements reoccur ''periodically'' throughout the table, hence the name 'periodic table'.
Ionization energy increases across the period hence the noble gas He has the highest ionization energy.
Since ionization energy increases across the period, group 1 elements possess the lowest ionization energy.
Since atomic size increases down the group and decreases across the period, gallium is smaller than indium, potassium is smaller than caesium, polonium is smaller than titanium and iodine is larger than bromine.
This explanation above justifies the order of increasing atomic radius of group 13 elements shown in answer number 8 above.
Since atomic size decreases across the period, the order of increasing atomic size shown in answer number 9 above is correct.
Ionization energy is the energy required to remove electrons from the outermost shell of an atom.
Answer:
It conducts electricity when it is dissolved in water.
Explanation:
It conducts electricity when it is dissolved in water is a property of a substance that is composed of atoms that are held together by ionic bonds.
<span>To find the percent composition of KCLO(2), calculate the total mass of the molecule: K has 39.0983 g/mol, Cl has 35.4532 g/mol, and O(2) is 31.99886 g/mol. This sums to 106.5501 g/mol. Next, find the percentages of each element: K would have (39.0983/106.5501) = 0.366948 or 36.6948%, Cl would have (35.4532/106.5501) = 0.332737 or 33.2737%, and O(2) would be (31.99886/106.5501) = 0.300318, or 30.0318%. Summing these three percentages gives 100.0003%, which is within the scope of a rounding error.</span>
Answer: pH of buffer solution is 8.1
Explanation:
The formula for the Henderson–Hasselbalch equation is:
![pH=pK_a+\log\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
is the concentration of ![[H^+]](https://tex.z-dn.net/?f=%5BH%5E%2B%5D)
is the acid dissociation constant,
and 
are concentrations of the conjugate base and starting acid.
Putting in the values we get:
Thus pH of buffer solution is 8.1
