That an ip leak don’t trust it ^^^^^^^^^^
The mass of 2.80 grams of h2o is 18.02 amu I believe
In a saturated solution, extra solid X would remain solid, dissolve in an unsaturated solution, and crystallize in a supersaturated one.
A solution is said to be saturated when there is a maximum amount of solute present that has been dissolved in the solvent. As a result, the system is in an equilibrium between the dissolved and undissolved solutes: A solution is considered to be unsaturated if the solute concentration is less than the equilibrium solubility. A supersaturated solution is one that has more solute than is necessary to generate a saturated solution at a given temperature.
Learn more about Supersaturated here-
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The safe house of lean might influence a product owner to provide better customer service.
<h3>What is the main purpose of safe house of lean?</h3>
The safe house of lean is designed to provide ultimate value to the customer one the shortest possible time.
Employing the safe house of lean might influence the product owner in the following two ways:
- enable the product ownerr work together with Agile Teams to develop stories with clear acceptance criteria.
- assist Agile Teams gain a clear understanding of their customers and how best to serve them.
Therefore, the safe house of lean might influence a product owner to provide better customer service.
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Answer:
Here's what I get
Explanation:
Assume the initial concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.
We must calculate the initial concentration of HI.
1. We will need a chemical equation with concentrations, so let's gather all the information in one place.
H₂ + I₂ ⇌ 2HI
I/mol·L⁻¹: 0.30 0.15 x
2. Calculate the concentration of HI
![Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} = 5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}](https://tex.z-dn.net/?f=Q_%7B%5Ctext%7Bc%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BHI%5D%7D%5E%7B2%7D%7D%20%7B%5Ctext%7B%5BH%24_%7B2%7D%24%5D%5BI%24_%7B2%7D%24%5D%7D%7D%20%3D%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.30%20%5Ctimes%200.15%7D%20%3D%20%205.56%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%200.30%20%5Ctimes%200.15%20%5Ctimes%205.56%20%3D%200.250%5C%5Cx%20%3D%20%5Csqrt%7B0.250%7D%20%3D%20%5Ctextbf%7B0.50%20mol%2FL%7D%5C%5C%5Ctext%7BThe%20initial%20concentration%20of%20HI%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.50%20mol%2FL%7D%7D%24%7D)
3. Plot the initial points
The graph below shows the initial concentrations plotted on the vertical axis.
