The correct answers are ,
A) C
B) N
C) Ti
D) Zn
E) Fe
F) Phosphorus
G)Calcium
H) Helium
I) Lead
J) Silver
<h3>How are elements named?</h3>
Elements have been given names based on a variety of factors, <u>including their characteristics</u>, the compound or ore from which they were extracted, the method by which they were found or acquired, mythical characters, locations, and well-known individuals. Some components have <u>names that are descriptive and are based on one of their attributes.</u>
The International Union of Pure and Applied Chemistry chooses the official element names and symbols (IUPAC). However, different nations frequently use similar names and symbols for elements. Official names and symbols for elements are not given until after their discovery has been confirmed. The discoverer may then suggest a name and a symbol.
There are name standards for several element groupings. Names of halogens end in -ine. All noble gas names, save helium, end in -on. The names of most other elements finish with -ium.
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Answer:
C. 0.20 M Mg ion & 0.40 M Cl ion
Explanation:
MgCl₂ is a ionic salt which is dissociated as this
MgCl₂ → Mg²⁺ + 2Cl⁻
First of all, we have a solution of 200 mL, with [MgCl₂] = 0.6M
Molarity . volume = moles.
0.6 mol/l . 0.2l = 0.12 mol
MgCl₂ → Mg²⁺ + 2Cl⁻
0.12mol 0.12 0.24
This moles are also in 400mL of water, so the new concentration is
[Mg²⁺] = 0.12 m/0.6L = 0.2M
[Cl⁻] = 0.24 m/0.6L = 0.4M
Remember we initially have 200mL and then, we add 400 mL, so we supose aditive volume. (600mL)
Answer:
see explanation
Explanation:
The process of ionization to produce cations is endothermic. For formation of Ca⁺² two ionization steps need be illustrated as follows...
1st ionization step: Ca° + 590Kj => Ca⁺ + e⁻
2nd ionization step: Ca⁺ + 1151Kj => Ca⁺² + e⁻
__________________________________-
Net Ionization Rxn: Ca° + 1741Kj => Ca⁺² + 2e⁻
Answer:
14 mL
Explanation:
To prepare a solution by a concentrated solution, we must use the equation:
C1xV1 = C2xV2, where <em>C</em> is the concentration, <em>V</em> is the volume, 1 is the initial solution and 2 the final solution.
The final solution must have 2 mL and a concentration of 350 pg/mL, and the initial solution has a concentration of 50 pg/mL.
Then:
50xV1 = 350x2
50xV1 = 700
V1 = 700/50
V1 = 14 mL
For the given molecule, we are asked to give-
- The electron configuration of an isolated B atom
- The electron configuration of an isolated F atom
- Hybrid orbitals should be constructed on the B atom to make the B–F bonds in Boron tri flouride
- valence orbitals, if any, remain unhybridized on the B atom.
- The electron configuration of an isolated B atom:
as atomic number of B is 5
electronic configuration will be [He] 2s² 2p¹
- The electron configuration of an isolated F atom:
as atomic number of F is 9
electronic configuration will be [He] 2s² 2p5
- Hybrid orbitals should be constructed on the B atom to make the B–F bonds in Boron tri flouride will be sp2.
as the one s and two of p orbital from the valance shell will hybridised to make 3 hybrid orbital of B resulting in 3 B-F bonds.
- valence orbitals, if any, remain unhybridized on the B atom will be 1
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