We know that the molarity of a solution is calculated by the following equation:

That being said we are given two of the 3 things that we need: Volume of water, as well as the molarity of the solution. Let's plug those in to the equation to find how many moles of

that we need:

=>

This means that we have 0.04 moles of

.
In order to calculate the number of grams that we need for this solution, we must first calculate how many grams of

are in 1 mole. We do this by taking the atomic masses of each element from the periodic table and adding them together.
Na = 22.99 g
S = 32.06 g
We have 2 Na's, and we have 1 S. So lets add them together:
2(22.99) + (32.06) =

.
Since we need 0.04 moles of

, we can multiply the molar weight of the molecule times the amount of moles needed to find the total grams that we need for the solution:

Now we know that in order to make a 0.2 M solution of

, we must use
3.1216 grams.
Answer:

Explanation:
Whenever a question asks you, "How long does it take to reach a certain concentration?" or something like that, you must use the appropriate integrated rate law expression.
The integrated rate law for a first-order reaction is
![\ln \left (\dfrac{[A]_{0}}{[A]} \right ) = kt](https://tex.z-dn.net/?f=%5Cln%20%5Cleft%20%28%5Cdfrac%7B%5BA%5D_%7B0%7D%7D%7B%5BA%5D%7D%20%5Cright%20%29%20%3D%20kt)
Data:
[A]₀ = 1.28 mol·L⁻¹
[A] = 0.17 [A]₀
k = 0.0632 s⁻¹
Calculation:
![\begin{array}{rcl}\ln \left (\dfrac{[A]_{0}}{0.170[A]_{0}} \right ) & = & 0.0632t\\\\\ln \left (5.882) & = & 0.0632t\\1.772 & = & 0.0632t\\\\t & = & \dfrac{1.772}{0.0632}\\\\t & = & \textbf{{28.0 s}}\\\end{array}\\\text{It will take } \boxed{\textbf{28.0 s}} \text{ for [HI] to decrease to 17.0 \% of its original value.}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5Cln%20%5Cleft%20%28%5Cdfrac%7B%5BA%5D_%7B0%7D%7D%7B0.170%5BA%5D_%7B0%7D%7D%20%5Cright%20%29%20%26%20%3D%20%26%200.0632t%5C%5C%5C%5C%5Cln%20%5Cleft%20%285.882%29%20%26%20%3D%20%26%200.0632t%5C%5C1.772%20%26%20%3D%20%26%200.0632t%5C%5C%5C%5Ct%20%26%20%3D%20%26%20%5Cdfrac%7B1.772%7D%7B0.0632%7D%5C%5C%5C%5Ct%20%26%20%3D%20%26%20%5Ctextbf%7B%7B28.0%20s%7D%7D%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7BIt%20will%20take%20%7D%20%5Cboxed%7B%5Ctextbf%7B28.0%20s%7D%7D%20%5Ctext%7B%20for%20%5BHI%5D%20to%20decrease%20to%2017.0%20%5C%25%20of%20its%20original%20value.%7D)
The mass of 40 mL of water is 40 grams. Since D = m/v and mL = cm3, the density of water is 1 g/cm3. Choose a volume between 1 and 100 mL. Use your graph to find the mass.
Quantitative observations include numerical data. Ex: 32 degrees, 10 inches, etc.