Answer:
The Sun is the closest star to Earth.
Part 1)
Cu- <span>[Ar] 3d¹⁰4s¹ </span><span>atomic number: 29
</span>
<span>O- [He] 2s2 2p<span>4 atomic number:8
</span></span>La- <span>[Xe] 5d¹ 6s² </span><span>atomic number:57
Y- </span><span>[Kr] 4d¹5s² </span><span>atomic number:39
Ba- </span><span>[Xe] 6s² </span><span>atomic number:56
Tl- </span><span>[Xe] 4f¹⁴ 5d¹⁰ 6s² 6p¹ </span><span>atomic number:81
Bi- </span> <span>[Xe] 4f¹⁴ 5d¹⁰ 6s² 6p³ </span>atomic number:83
Part 2)
You are able to this by consulting the periodic table and following this steps:
-Find your atom's atomic number;
<span>-Determine the charge of the atom (these were all uncharged)
</span><span>-Memorize the order of orbitals (s, d, p, d.. and how many electrons they can fit)
</span>-<span>Fill in the orbitals according to the number of electrons in the atom
- </span><span>for long electron configurations, abbreviate with the noble gases</span>
Question 22: conclusion
Question 23: analyze
Answer:
The answers are in the explanation
Explanation:
A buffer is the mixture of a weak acid with its conjugate base or vice versa. Thus:
<em>1)</em> Mixing 100.0 mL of 0.1 M HF with 100.0 mL of 0.05 M mol KF. <em>Will </em>result in a buffer because HF is a weak acid and KF is its conjugate base.
<em>2)</em> Mixing 100.0 mL of 0.1 M NH₃ with 100.0 mL of 0.1 M NH₄Br. <em>Will not </em>result in a buffer because NH₃ is a strong base.
<em>3) </em>Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.05 M KOH. <em>Will </em>result in a buffer because HCN is a weak acid and its reaction with KOH will produce CN⁻ that is its conjugate base.
<em>4)</em> Mixing 100.0 mL of 0.1 M HCl with 100.0 mL of 0.1 M KCl <em>Will not </em>result in a buffer because HCl is a strong acid.
<em>5)</em> Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.1 M KOH <em>Will not </em>result in a buffer because each HCN will react with KOH producing CN⁻, that means that you will have just CN⁻ (Conjugate base) without HCN (Weak acid).
I hope it helps!
<u>Answer:</u> The wavelength of the flame is 462 nm and color of cesium flame is blue.
<u>Explanation:</u>
To calculate the wavelength, we use Planck's equation, which is:
where,
E = Energy of 1 photon =
h = Planck's constant =
c = speed of light =
= wavelength = ?
Putting values in above equation, we get:
The range of wavelength of blue light lies in range of 500 nm - 435 nm
The calculated wavelength lies in the above range. So, the color of the cesium flame is 462 nm
Hence, the wavelength of the flame is 462 nm and color of cesium flame is blue.