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MatroZZZ [7]
3 years ago
9

How many shells does Potassium (K) have?​

Chemistry
2 answers:
slamgirl [31]3 years ago
6 0

Answer:19 i think because if Potassium atoms have 19 electrons and 19 protons with one valence electron in the outer shell.

Airida [17]3 years ago
4 0

Answer:

Potassium. Potassium is the fourth element in the first column of the periodic table. It is classified as an alkali metal. Potassium atoms have 19 electrons and 19 protons with one valence electron in the outer shell.

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Molar mass for C 3 H 6 (O H)2
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Answer:

The answer is 76.0944

Explanation:

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ion

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3 years ago
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Does ice particles have high interparticle force of attraction? Justify your answer.<br> please and
Schach [20]

Answer:

¿Las partículas de hielo tienen una alta fuerza de atracción entre partículas? Justifica tu respuesta

Explanation:

Dichos contenidos están presentes en los currículos de Física y Química de la educación básica, con independencia del marco legal, pues introducen al alumno en el conocimiento químico de la materia. Aunque la teoría cinética molecular obvia la composición atómica de las partículas, no deja de ser un contenido deseable para introducir a los alumnos en el mundo de la química pues permite diferenciar y establecer relaciones entre los niveles macro, micro y simbólico de la materia.

8 0
3 years ago
What is the energy of a 933 nm wave?
vovikov84 [41]

E=hc/l

E=

<span><span>E=<span>(6.626 x 10-34 J s)(3.0 x 108m/s )</span><span>=2.88 x 10-19J</span></span><span>6.90 x 10-7m</span></span>
8 0
3 years ago
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EDTA EDTA is a hexaprotic system with the p K a pKa values: p K a1 = 0.00 pKa1=0.00 , p K a2 = 1.50 pKa2=1.50 , p K a3 = 2.00 pK
mihalych1998 [28]

Answer:

Check the explanation

Explanation:

When,

pH = -log[H+] = 3.30

[H+] = 5.0 X 10^{-4} M

Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 7.4 X 10^{-7} ; Ka6 = 4.3 X 10^{-11}

alpha[Y^-4] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6

= 1.56 X 10^{-20} + 3.12 X 10^{-17} + 2 X 10^{-15} + 4 X 10^{-14} + 1.6 X 10^{-13} + 2.34 X 10^{-16} + 2 X 10^{-23}

= 2.02 X 10^{-13}

When,

pH = -log[H+] = 10.15

[H+] = 7.08 X 10^{-11} M

Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 7.4 X 10^{-7} ; Ka6 = 4.3 X 10^-11

alpha[Y^{-4}] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6

= 1.26 X 10^{-61} + 1.8 X 10^{-51} + 8.1 X 10^{-43} + 1.12 X 10^{-34} + 3.17 X 10^{-27} + 3.3 X 10^{-23} + 1.83 X 10^{-23}

= 5.12 X 10^{-23}

4 0
3 years ago
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