The concentration of the chemist's working solution is 6.718 mol/L.
Given,
The volume of silver perchlorate solution (V₁) = 790 ml.
The volume of distilled water added to the silver perchlorate(V₂) = 290ml.
The concentration of a stock solution of silver perchlorate (C₂) = 18.3 mol/L.
We have to find the concentration of the chemist's working solution (C₁).
As we know,
V₁C₁ = V₂C₂
From the above formula, we get,
C₁ = V₂C₂/V₁
C₁ = 290 × 18.3/ 790
C₁ = 6.718 mol/L
The concentration of the working solution = 6.718 mol/L
Hence, the concentration of the chemist's working solution is 6.718 mol/L.
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Answer:

Explanation:
Whenever a question asks you, "What is the concentration after a given time?" or something like that, you must use the appropriate integrated rate law expression.
The reaction is 2nd order, because the units of k are L·mol⁻¹s⁻¹.
The integrated rate law for a second-order reaction is
![\dfrac{1}{\text{[A]}} =\dfrac{1}{\text{[A]}_{0}}+ kt](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B%5Ctext%7B%5BA%5D%7D%7D%20%3D%5Cdfrac%7B1%7D%7B%5Ctext%7B%5BA%5D%7D_%7B0%7D%7D%2B%20kt)
Data:
k = 2.4 × 10⁻²¹ L·mol⁻¹s⁻¹
[A]₀ = 0.0100 mol·L⁻¹
[A] = 0.009 00 mol·L⁻¹
Calculation
:
![\begin{array}{rcl}\dfrac{1}{\text{[A]}} & = & \dfrac{1}{\text{[A]}_{0}}+ kt\\\\\dfrac{1}{0.00900 }& = & \dfrac{1}{0.0100} + 2.4 \times 10^{-21} \, t\\\\111.1&=& 100.0 + 2.4 \times 10^{-21} \, t\\\\11.1& = & 2.4 \times 10^{-21} \, t\\t & = & \dfrac{11.1}{ 2.4 \times 10^{-21}}\\\\& = & \mathbf{4.6 \times 10^{21}}\textbf{ s}\\\end{array}\\\text{It will take $\large \boxed{\mathbf{4.6 \times 10^{21}}\textbf{ s}}$ for the HI to decompose}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5Cdfrac%7B1%7D%7B%5Ctext%7B%5BA%5D%7D%7D%20%26%20%3D%20%26%20%5Cdfrac%7B1%7D%7B%5Ctext%7B%5BA%5D%7D_%7B0%7D%7D%2B%20kt%5C%5C%5C%5C%5Cdfrac%7B1%7D%7B0.00900%20%7D%26%20%3D%20%26%20%5Cdfrac%7B1%7D%7B0.0100%7D%20%2B%202.4%20%5Ctimes%2010%5E%7B-21%7D%20%5C%2C%20t%5C%5C%5C%5C111.1%26%3D%26%20100.0%20%2B%202.4%20%5Ctimes%2010%5E%7B-21%7D%20%5C%2C%20t%5C%5C%5C%5C11.1%26%20%3D%20%26%202.4%20%5Ctimes%2010%5E%7B-21%7D%20%5C%2C%20t%5C%5Ct%20%26%20%3D%20%26%20%5Cdfrac%7B11.1%7D%7B%202.4%20%5Ctimes%2010%5E%7B-21%7D%7D%5C%5C%5C%5C%26%20%3D%20%26%20%5Cmathbf%7B4.6%20%5Ctimes%2010%5E%7B21%7D%7D%5Ctextbf%7B%20s%7D%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7BIt%20will%20take%20%24%5Clarge%20%5Cboxed%7B%5Cmathbf%7B4.6%20%5Ctimes%2010%5E%7B21%7D%7D%5Ctextbf%7B%20s%7D%7D%24%20for%20the%20HI%20to%20decompose%7D)
When a light source moves away for an observer.
There are two problems for this question:1. What is the total dollar amount of your profit and loss:
Put option premium is equal to 0.04 per unit.
The exercise price is 1.22
One option contract is 100,000
Selling price is 1.20
-Purchase prise is - 1.22
-Premium paid is +0.04
Net profit is = 0.02 x 100,000 = 2,000 – 80 = 1,920
2. Now undertake that as an alternative of taking a position in the put option one year ago, you sold a future's contract on 100,000 euros with a payment date of one year.
Find the total dollar amount of your profit or loss.
Solution: Contract to buy: $1.20 x 100,000 = 120,000 at payment date.
Contract to sell: $1.22 x 100,000 = 122,000 at settlement date
Settle contracts: -2,000 - 80 = -$2,080