From a Chemistry perspective it is the electron at 1/1800th the mass of a proton or neutron.
Answer:
Lithium,Sodium,Potassium all belong to transition metals
<u>Answer:</u> Copper (I) iodide will precipitate first.
<u>Explanation:</u>
We are given:
of CuCl = 
of CuI = 
Concentration of 
Concentration of 
Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
![K_{sp}=[Cu^+][Cl^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCu%5E%2B%5D%5BCl%5E-%5D)
Putting values in above equation, we get:
![1.0\times 10^{-6}=[Cu^+]\times 0.021](https://tex.z-dn.net/?f=1.0%5Ctimes%2010%5E%7B-6%7D%3D%5BCu%5E%2B%5D%5Ctimes%200.021)
![[Cu^+]=\frac{1.0\times 10^{-6}}{0.021}=4.76\times 10^{-5}M](https://tex.z-dn.net/?f=%5BCu%5E%2B%5D%3D%5Cfrac%7B1.0%5Ctimes%2010%5E%7B-6%7D%7D%7B0.021%7D%3D4.76%5Ctimes%2010%5E%7B-5%7DM)
Concentration of copper (I) ion = 
![K_{sp}=[Cu^+][I^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCu%5E%2B%5D%5BI%5E-%5D)
Putting values in above equation, we get:
![5.1\times 10^{-12}=[Cu^+]\times 0.017](https://tex.z-dn.net/?f=5.1%5Ctimes%2010%5E%7B-12%7D%3D%5BCu%5E%2B%5D%5Ctimes%200.017)
![[Cu^+]=\frac{5.1\times 10^{-12}}{0.017}=3.00\times 10^{-10}M](https://tex.z-dn.net/?f=%5BCu%5E%2B%5D%3D%5Cfrac%7B5.1%5Ctimes%2010%5E%7B-12%7D%7D%7B0.017%7D%3D3.00%5Ctimes%2010%5E%7B-10%7DM)
Concentration of copper (I) ion = 
For the precipitation of copper (I) ions, we need less concentration of copper (I) ions. So, copper (I) iodide will precipitate first.
Answer:
92.9%
Explanation:
You have been given the actual yield of the reaction. First, you need to find the theoretical yield of the reaction. To do this, you need to (1) convert grams Fe₂O₃ to moles Fe₂O₃ (via molar mass from periodic table values), then (2) convert moles Fe₂O₃ to moles Fe (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles Fe to grams Fe (via molar mass).
Once you have found the theoretical yield, you need to use the percent yield equation to calculate the final answer. This number should have 3 sig figs to match the given values.
<u>(Step 1)</u>
Molar Mass (Fe₂O₃): 2(55.845 g/mol) + 3(15.998 g/mol)
Molar Mass (Fe₂O₃): 159.684 g/mol
1 Fe₂O₃(s) + 3 CO(g) ---> 2 Fe(s) + 3 CO₂(g)
Molar Mass (Fe): 55.845 g/mol
50.0 g Fe₂O₃ 1 mole 2 moles Fe 55.845 g
-------------------- x ------------------ x --------------------- x ---------------- = 35.0 g Fe
159.684 g 1 mole Fe₂O₃ 1 mole
<u>(Step 2)</u>
Actual Yield
Percent Yield = --------------------------- x 100%
Theoretical Yield
32.5 g Fe
Percent Yield = ---------------------- x 100% = 92.9%
35.0 g Fe
Answer:
n = 2
Explanation:
We will solve this problem using the Rydberg equation since all the information is given. One has to be very careful in solving for the energy level becuse the numbers are slightly messy as is solving for n.
When the energy of the transition is given in wavelength it is best to use the Rydberg equation in the form:
1/Lambda = Rh x Z² x ( 1/ n₁² - 1/ n₂²)
Lambda= 30.4 x 10 ⁻⁹ m ( 1nm = 1 x 10⁻⁹ m)
Rh = 10973700 m⁻¹
Z = 4 for Be³⁺
1/(30.4 x 10⁻⁹ )= 10973700 x (16) x (1/ n² - 1/4²)=
0.1873 + 1/16 = 1/ n²
n² = 1/ 0.25
n ² = 4
n= √4 = 2
