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)
Answer:
electron-electron repulsion
Explanation:
When electrons add into valence shell of neutral elements, the element assumes a negative oxidation state. With this, the number of electrons having (-) charges will be larger than the number of protons having positive (+) charges. As a result, the extra electrons repel one another (i.e., like charges repel) and a larger radius is the result.
In contrast, when cations are formed, electrons are removed from the valence level (oxidation) producing an element having a greater number of protons than electrons. The larger number of protons will function to attract the electron cloud with a greater force that results in a contraction of atomic radius and a smaller spherical volume than the neutral unionized element.
To visualize, see attached chart that shows atomic and ionic radii before and after ionization of the elements.
Answer:
0.295 L
Explanation:
It seems your question lacks the final concentration value. But an internet search tells me this might be the complete question:
" A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to 24.0 mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits. "
Keep in mind that if your value is different, the answer will be different as well. However the methodology will remain the same.
To solve this problem we can<u> use the formula</u> C₁V₁=C₂V₂
Where the subscript 1 refers to the concentrated solution and the subscript 2 to the diluted one.
- 47.2 mL * 150 mM = 24.0 mM * V₂
And <u>converting into L </u>becomes:
- 295 mL *
= 0.295 L
Hard water contains high levels of minerals such as magnesium and calcium and can be a real pain to remove if left untreated. When hard water is boiled in an electric kettle, these minerals do not evaporate like the water does but instead remains in the kettle.
Answer: 2,1,1
Explanation: (2)Na2HPO4=(1)Na4P2O7+(1)H20
