Bromine has the following electron configuration: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5. categorize the electrons in each. Answer for video: The video player is loaded.
On the periodic chart, row 5, column 7, is where you can find a chemical element that was identified in 1811. It has a proton count of 53 and an atomic mass of 126.9. Iodine's atom, then, contains 53 electrons in the following configuration: 1s2, 2s2, 2p6, 3s2, 3d10, 4p6, 5s2, 4d10, 5p5 (Kr 4d10 5s2 5p5). Cu Z = 29 has an electrical arrangement of 1s2 2s2 2p6 3s2 3p6 3d10 4s1. Copper (Co) has the following electron configuration: 1s2 2s2 2p6 3s3 3p6 4s2 3d7. If a chemist were to refer to Copper by its subshell, they would abbreviate this notation to "3d7."
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<span>Answer is: the mass of hydrogen is 22,05 grams.
m(</span>Al(C₂H₃O₂)₃)<span> = 500 g.
M</span>(Al(C₂H₃O₂)₃) = 27 + 6 ·12 + 9 · 1 + 6 · 16 · g/mol = 204 g/mol.<span>
n</span>(Al(C₂H₃O₂)₃) = m(Al(C₂H₃O₂)₃) ÷ M(Al(C₂H₃O₂)₃).
n(Al(C₂H₃O₂)₃) = 500 g ÷ 204 g/mol.
n(Al(C₂H₃O₂)₃) = 2,45 mol.
n(Al(C₂H₃O₂)₃) : n(H) = 1 : 9.
n(H) = 22,05 mol.
m(H) = 22,05 mol · 1 g/mol
m(H) = 22,05 g.
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The question is incomplete, here is the complete question:
What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)
<u>Answer:</u> The volume of HCl neutralized is 1.25 mL
<u>Explanation:</u>
To calculate the volume of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of stomach acid which is HCl
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
Hence, the volume of HCl neutralized is 1.25 mL
