A solution has an absorbance of 0.2 with a path length of 1 cm. Given the molar absorptivity coefficient is 59 cm⁻¹ M⁻¹, the molarity is 0.003 M.
<h3>What does Beer-Lambert law state?</h3>
The Beer-Lambert law states that for a given material sample, path length and concentration of the sample are directly proportional to the absorbance of the light.
A solution has an absorbance of 0.2 with a path length of 1 cm. Given the molar absorptivity coefficient is 59 cm⁻¹ M⁻¹, we can calculate the molarity of the solution using the following expression.
A = ε × b × c
c = A / ε × b
c = 0.2 / (59 cm⁻¹ M⁻¹) × 1 cm = 0.003 M
where,
- A is the absorbance.
- ε is the path length.
- b is the molar absorptivity coefficient.
- c is the molar concentration.
A solution has an absorbance of 0.2 with a path length of 1 cm. Given the molar absorptivity coefficient is 59 cm⁻¹ M⁻¹, the molarity is 0.003 M.
Learn more about the Beer-Lambert law here: brainly.com/question/12975133
Answer:
60g
Explanation:
Moles = mass / mollar mass
Mass = Moles x mollar mass
Mass = 2.50 x 24
Mass = 60g
For this question, assume that you have 1 compound. This compound is divided in half once, so you are left with 0.5. That 0.5 that remains is divided in half again, this is the second half-life, and you are left with 0.25. The final half life involves dividing 0.25 in half, which means you are left with 0.125. For the answer to make sense, you need to know your conversions between decimals and fractions. To make it simple, if you have 0.125 and you times it by 8, you are left with your initial value of 1. Therefore, after three half-lives, you are left with 1/8th of the compound.
Answer : The balanced reaction in acidic solution is,
Explanation :
The given partial equation is,
First we have to separate into half reaction. The two half reactions are:
Now we have to balance the half reactions in acidic medium, we get:
............(1)
............(2)
Now we have to balance the electrons of the half reactions. When we are multiplying the equation (1) by 2, we get
...........(3)
Now we have to add both the half reactions (2) and (3), we get the final balanced chemical reaction.
