Answer:
0.500 mole of Xe (g) occupies 11.2 L at STP.
General Formulas and Concepts:
<u>Gas Laws</u>
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
<u>Stoichiometry</u>
- Mole ratio
- Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
<em>Identify.</em>
0.500 mole Xe (g)
<u>Step 2: Convert</u>
- [DA] Set up:
- [DA] Evaluate:
Topic: AP Chemistry
Unit: Stoichiometry
Given the data from the question, the mass of arsenic that contains 1.23×10²⁰ atoms is 0.0153 g
<h3>Avogadro's hypothesis </h3>
6.02×10²³ atoms = 1 mole of arsenic
But
1 mole of arsenic = 75 g
Thus, we can say that:
6.02×10²³ atoms = 75 g of arsenic
<h3>How to determine the mass that contains 1.23×10²⁰ atoms</h3>
6.02×10²³ atoms = 75 g of arsenic
Therefore,
1.23×10²⁰ atoms = (1.23×10²⁰ × 75) / 6.02×10²³ atoms)
1.23×10²⁰ atoms = 0.0153 g of arsenic
Thus, 1.23×10²⁰ atoms is present in 0.0153 g of arsenic
Learn more about Avogadro's number:
brainly.com/question/26141731
Answer:
Explanation:
For a first order reaction the rate law is:
![v=\frac{-d[A]}{[A]}=k[A]](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B-d%5BA%5D%7D%7B%5BA%5D%7D%3Dk%5BA%5D)
Integranting both sides of the equation we get:
![\int\limits^a_b {\frac{d[A]}{[A]}} \, dx =-k\int\limits^t_0 {} \, dt](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7B%5Cfrac%7Bd%5BA%5D%7D%7B%5BA%5D%7D%7D%20%5C%2C%20dx%20%3D-k%5Cint%5Climits%5Et_0%20%7B%7D%20%5C%2C%20dt)
where "a" stands for [A] (molar concentration of a given reagent) and "b" is {A]0 (initial molar concentration of a given reagent), "t" is the time in seconds.
From that integral we get the integrated rate law:
![ln\frac{[A]}{[A]_{0} } =-kt](https://tex.z-dn.net/?f=ln%5Cfrac%7B%5BA%5D%7D%7B%5BA%5D_%7B0%7D%20%7D%20%3D-kt)
![[A]=[A]_{0}e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA%5D_%7B0%7De%5E%7B-kt%7D)
![ln[A]=ln[A]_{0} -kt](https://tex.z-dn.net/?f=ln%5BA%5D%3Dln%5BA%5D_%7B0%7D%20-kt)
![k=\frac{ln[A]_{0}-ln[A]}{t}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7Bln%5BA%5D_%7B0%7D-ln%5BA%5D%7D%7Bt%7D)
therefore k is
One would be phosporous whose configuration is 1s2 2s2 2p6 3s2 3p3
Answer:
0.136g
Explanation:
A student dissolved 5.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution and then diluted it with water to give 275. mL of a final solution. How many grams of NO3- ion are there in the final solution?
Initial mole of Co(NO3)2 
Mole of Co(NO3)2 in final solution
Mole of NO3- in final solution = 2 x Mole of Co(NO3)2
Mass of NO3- in final solution is mole x Molar mass of NO3
