Answer:
S = 7.9 × 10⁻⁵ M
S' = 2.6 × 10⁻⁷ M
Explanation:
To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0
C +S +S
E S S
The solubility product (Ksp) is:
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²
S = 7.9 × 10⁻⁵ M
<u>Solubility in 0.0120 M CoBr₂ (S')</u>
First, we will consider the ionization of CoBr₂, a strong electrolyte.
CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)
1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.
Then,
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0.0240
C +S' +S'
E S' 0.0240 + S'
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')
In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.
S' = 2.6 × 10⁻⁷ M
Answer:
There is 54.29 % sample left after 12.6 days
Explanation:
Step 1: Data given
Half life time = 14.3 days
Time left = 12.6 days
Suppose the original amount is 100.00 grams
Step 2: Calculate the percentage left
X = 100 / 2^n
⇒ with X = The amount of sample after 12.6 days
⇒ with n = (time passed / half-life time) = (12.6/14.3)
X = 100 / 2^(12.6/14.3)
X = 54.29
There is 54.29 % sample left after 12.6 days
Answer:
According to Boyle's Law, an inverse relationship exists between pressure and volume. ... The relationship for Boyle's Law can be expressed as follows: P1V1 = P2V2, where P1 and V1 are the initial pressure and volume values, and P2 and V2 are the values of the pressure and volume of the gas after change.
<em>Hope that helps! :)</em>
<em></em>
<em>-Aphrodite</em>
Explanation:
Answer:
The answer to your question is below
Explanation:
1. Found in period 2. All the elements in the list are found in period 2.
a. F This option is correct
b. Be Beryllium is located in period two.
c. O also oxygen is found in period 2.
d. C Carbon is found in period 2.
2.- Can gain lose 4 electrons to become its nearest stable noble gas. Only Carbon.
a. F This option is wrong, F becomes stable when it gains 1 electron.
b. Be Beryllium becomes stable when it loses 2 electrons.
c. O Become stable when it gains 2 electrons.
<u>d. C </u><u>Become stable when it gains or loses 4 electrons.</u>
Explanation:
Since, it is given that critical temperature of Argon is 150.9 K and critical pressure of Argon is 48.0 atm.
It is known that gas phase of neon occurs at 50 K. As the boiling point of Ar is more than the boiling point of neon which means that there is strong intermolecular force of attraction between argon molecules as compared to neon molecules.
This is also because argon is larger in size. As a result, induced dipole-induced dipole forces leads to more strength in Ar as compared to Ne.
