Answer:
The correct answer is 5.6 × 10⁻²³ M.
Explanation:
As a highly soluble salt, KBr dissolves easily in water, while Hg₂Br₂ is very less soluble in comparison to KBr.
Let the solubility of Hg₂Br₂ is S mol per liter.
Therefore,
KBr (s) (1.0 M) ⇒ K⁺ (aq) (1M) + Br⁻ (aq) (1M)
Hg₂Br₂ (s) (1-S) ⇔ Hg₂⁺ (aq) (S) + 2Br⁻ (aq) (2S)
Net [Br-] = (2S + 1) M
Ksp = S (2S + 1)²
Ksp = S (4S² + 1 + 4S)
Ksp = 4S³ + S + 4S²
As the solubility is extremely less, therefore, we can ignore S² and S³. Now,
Ksp = S = 5.6 × 10⁻²³ M
Hence, the solubility of Hg₂Br₂ is 5.6 × 10⁻²³ M.
Answer:
Marcie pick up a rock
Explanation:
A rock is a natural occuring no living thing that has a definite structure from the physical characteristics Marcie saw in the rock she picked it tends to rock because unlike minerals rocks has definite structure.
V=650 ml = 0.65 l
c=0.4 mol/l
n=vc
n=0.65 l * 0.4 mol/l = 0.26 mol
The only solution you'll need to have for this problem is a periodic table. The columns in the table are called groups, and they are number from the left to the right starting with 1. The rows in the table are called periods which are numbered from the top to bottom starting with 1.
6. Elements that belong to the same group portray similar chemical properties. Therefore, the element in period 4 which is also in group 2 is
<em>Calcium (Ca)</em>.
7. The elements that are striked through with the red slanting lines are the metalloids. All the elements to the left of the metalloids are metals. All the elements to the right are nonmetals. Bromine has a symbol of Br. Since At is a metalloid and located in the same group with Br, the
<em>answer is Astatine (At).</em>8. Tin has the chemical symbol of Sn. The nonmetal that is located in the same group is
<em>Carbon (C)</em>.
9. All the elements in period 6 would have similar properties. The answers could be:
<em>Phosphorus (P), Arsenic (As), Antimony (Sb) and Bismuth (Bi)</em>.
10. Period is row 1 and group 18 is the last column.
<em>The answer is Helium (He).</em>
Lyman Series Working Formula:
1/λ = RH (1-(1/n^2))
Given:
n = 6
RH = Rydberg's constant = 1.0968x10^7 m^-1
c = speed of light = 3x10^8 m/s
Required:
Frequency (Hertz or cycles per second)
Solution:
To solve for the wavelength λ, we substitute the given in the working formula
1/λ = RH (1-(1/n^2))
1/λ = 1.0968x10^7 m^-1 (1-(1/6^2))
λ = 0.0000000938 m or 93.8 nm
To get the frequency, we will use the formula below.
f = c/λ
We then substitute c or the speed of light,
f = (3x10^8 m/s) / 0.0000000938 m
Therefore,
f = 3.2x10^15 s^-1
<em>ANSWER: Frequency = </em><em /><em>3.2x10^15 s^-1</em>