Answer:
- Volume = <u>2.0 liter</u> of 1.5 M solution of KOH
Explanation:
<u>1) Data:</u>
a) Solution: KOH
b) M = 1.5 M
c) n = 3.0 mol
d) V = ?
<u>2) Formula:</u>
Molarity is a unit of concentration, defined as number of moles of solute per liter of solution:
<u>3) Calculations:</u>
- Solve for n: M = n / V ⇒ V = n / M
- Substitute values: V = 3.0 mol / 1.5 M = 2.0 liter
You must use 2 significant figures in your answer: <u>2.0 liter.</u>
Answer:
250 mL (total solution) = 104 mL (stock solution) + 146 mL (water)
Explanation:
Data Given
M1 = 6.00 M
M2 = 2.5 M
V1 = 250 mL
V2 = ?
Solution:
As the chemist needs to prepare 250 mL of solution from 6.00 M ammonium hydroxide solution to prepare a 2.50 M aqueous solution of ammonium hydroxide.
Now
first he have to determine the amount of ammonium hydroxide solution that will be taken from6.00 M ammonium hydroxide solution
For this Purpose we use the following formula
M1V1=M2V2
Put values from given data in the formula
6 x V1 = 2.5 x 250
Rearrange the equation
V1 = 2.5 x 250 /6
V1 = 104 mL
So 104 mL is the volume of the solution which we have to take from the 6.00 M ammonium hydroxide solution to prepare 2.5 M aqueous solution of ammonium hydroxide
But we have to prepare 250 mL of the solution.
so the chemist will take 104 mL from 6.00 M ammonium hydroxide solution and have to add 146 mL water to make 250 mL of new solution.
in this question you have to tell about the amount of water that is 146 mL
250 mL (total solution) = 104 mL (stock solution) + 146 mL (water)
Answer:
The concentration of the murexide solution is 0.0000745 M
Explanation:
From Beer-Lambert's law,
A = εlc
A = Absorbance = 28.65% = 0.2865
ε = molar absorptivity = 3847 M/cm
l = path length = 1cm
c = concentration in mol/L = ?
c = A/εl = 0.2865/(3847×1) = 0.0000745 mol/L
Hope this Helps!
Answer:
Energy lost is 7.63×10⁻²⁰J
Explanation:
Hello,
I think what the question is requesting is to calculate the energy difference when an excited electron drops from N = 15 to N = 5
E = hc/λ(1/n₂² - 1/n₁²)
n₁ = 15
n₂ = 5
hc/λ = 2.18×10⁻¹⁸J (according to the data)
E = 2.18×10⁻¹⁸ (1/n₂² - 1/n₁²)
E = 2.18×10⁻¹⁸ (1/15² - 1/5²)
E = 2.18×10⁻¹⁸ ×(-0.035)
E = -7.63×10⁻²⁰J
The energy lost is 7.63×10⁻²⁰J
Note : energy is lost / given off when the excited electron jumps from a higher energy level to a lower energy level
