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ziro4ka [17]
3 years ago
7

Explain the four quantum numbers

Chemistry
1 answer:
devlian [24]3 years ago
7 0

 Principal (shell) quantum number - n 
 

<span>Describes the energy level in the atom (1 through 7)<span>The maximum number of electrons in n is <span>2n2</span></span></span><span>  </span>
<span>2. Angular (subshell) quantum number - </span>l 
 <span><span>Describes the sublevel in n</span><span>Each energy level has n sublevels</span></span><span>  </span>
<span>3. Magnetic quantum number - </span>m 
 <span><span>Describes the orbital within l</span>s has 1 orbitalp has 3 orbitalsd has 5 orbitalsf has 7 orbitals</span><span>  </span>
<span>4. Spin quantum number - </span>s 
 <span>Describes the spin of the electrons in an orbital.Electrons in the same orbital must have oppisite spins.<span>Possible spins are clockwise or counterclockwise.</span></span>
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Write the overall, balanced molecular equation and indicate which element is oxidized and which is reduced for the following rea
galina1969 [7]

Answer:

Cd(s) + AgNO₃(aq)  → Cd(NO₃)₂ (aq) + Ag(s)

Oxidized: Cd

Reduced: Ag

Explanation:

Cd(s) + AgNO₃(aq)  → Cd(NO₃)₂ (aq) + Ag(s)

Cd → Cd²⁺  +  2e⁻      Half reaction oxidation

1e⁻ + Ag⁺ → Ag           Half reaction reduction

Ag changed oxidation number from +1 to 0

Cd changed oxidation number from 0 to +2

Let's ballance the electrons

( Cd → Cd²⁺  +  2e⁻ ) .1

( 1e⁻ + Ag⁺ → Ag ) .2

Cd + 2e⁻ + 2Ag⁺  → 2Ag +  Cd²⁺  +  2e⁻

Finally the ballance equation is:

Cd(s) + 2AgNO₃(aq)  → Cd(NO₃)₂ (aq) + 2Ag(s)

4 0
3 years ago
Flourine is found to undergo 10% radioactivity decay in 366 minutes determine its halflife​
yuradex [85]

Answer:

\boxed{\text{2408 min}}

Explanation:

The integrated rate law for radioactive decay is

\ln\dfrac{N_{0}}{N_{t}} = kt

1. Calculate the decay constant

\begin{array}{rcl}\ln \dfrac{100}{90} & = & k \times 366\\\\1.054 & = & 366k\\\\k & = & \dfrac{1.054 }{366}\\\\k & = & 2.879 \times 10^{-4} \text{ min}^{-1}\\\end{array}\\\\

2. Calculate the half-life

t_{\frac{1}{2}} = \dfrac{\ln2}{k}\\\\t_{\frac{1}{2}} = \dfrac{\ln2}{2.879 \times 10^{-4} \text{ min}^{-1}} = \text{2408 min}\\\\\text{The half-life for decay is } \boxed{\textbf{2408 min}}

8 0
3 years ago
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