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Slav-nsk [51]
3 years ago
5

PLS HELP ASAP!!!! Describe metallic bonds, include an explanation of their valence electrons. How this related to the properties

of metals?
Chemistry
1 answer:
Aloiza [94]3 years ago
5 0

If you look it up it will give you plenty of information. This is what I found:

The valence electrons of metals move freely in this way because metals have relatively low electronegativity, or attraction to electrons. The positive metal ions form a lattice-like structure held together by all the metallic bonds. ... When nonmetals bond together, the atoms share valence electrons and do not become ions

https://www.ck12.org/c/physical-science/metallic-bond/lesson/Metallic-Bonding-MS-PS/

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Sloan [31]

Answer:

last one

Explanation:

3 0
1 year ago
An atom of calcium loses two electrons. What is the charge on an ion of calcium?
kap26 [50]
The charge of an ion of calcium that loses two electrons is  D.) +2

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4 0
3 years ago
How to I find the number of electrons in and Atom?
Inessa [10]
Look at the atomic number of an element on the periodic table which is the smaller number.
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8 0
3 years ago
2AlF3 + 3K2O → 6KF + Al2O3<br><br> How many grams of AlF3would it take to make 15.524 g of KF?
mr_godi [17]
<h3>Answer:</h3>

7.4797 g AlF₃

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Stoichiometry</u>

  • Reading a Periodic Table
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN] 2AlF₃ + 3K₂O → 6KF + Al₂O₃

[Given] 15.524 g KF

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol AlF₃ = 6 mol KF

Molar Mass of Al - 26.98 g/mol

Molar Mass of F - 19.00 g/mol

Molar Mass of K - 39.10 g/mol

Molar Mass of AlF₃ - 26.98 + 3(19.00) = 83.98 g/mol

Molar Mass of KF - 39.10 + 19.00 = 58.10 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up:                              \displaystyle 15.524 \ g \ KF(\frac{1 \ mol \ KF}{58.10 \ g \ KF})(\frac{2 \ mol \ AlF_3}{6 \ mol \ KF})(\frac{83.98 \ g \ AlF_3}{1 \ mol \ AlF_3})
  2. Multiply/Divide:                                                                                                    \displaystyle 7.47966 \ g \ AlF_3

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 5 sig figs.</em>

7.47966 g AlF₃ ≈ 7.4797 g AlF₃

3 0
2 years ago
Read 2 more answers
Ca(OH)2(aq)+HNO3(aq)→H2O(l)+Ca(NO3)2(aq)
shtirl [24]

what are we doing in that question

Explanation:

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