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iris [78.8K]
3 years ago
15

Which type of substances have chemical bonds that are not directional and valence electrons that move freely between the atoms?

Chemistry
2 answers:
Wittaler [7]3 years ago
5 0
I believe the correct answer from the choices listed above is option C. The type of substances that has <span>chemical bonds that are not directional and valence electrons that move freely between the atoms are metals. Hope this answers the question. Have a nice day.</span>
Ket [755]3 years ago
5 0

Answer:

c) metals

Explanation:

In metallic bonding there are delocalized electrons that are shared by a number of metal cations. The electrons are free to move and are not confined to a single metal cation. These free flowing electrons give metals its specific properties such as ductility, malleability, electrical conductivity and more.

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pav-90 [236]
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7 0
3 years ago
Calculate the pH of 0.50M H2S
-BARSIC- [3]
I got that pH=3.65 using the fact that Ka=[H⁺][A⁻]/[HA] at equilibrium.  In the ice table, I stands for initial, C stands for change, and E stands for equilibrium.  

I hope this helps.  Let me know if anything is unclear.

5 0
3 years ago
Is centigrams or grams bigger.<br> example: which unit is larger.<br> 35cg or 35 g
sp2606 [1]

Answer:

35g

Explanation:

5 0
3 years ago
Read 2 more answers
When 74.8g of alanine C3H7NO2 are dissolved in 1450.g of a certain mystery liquid X, the freezing point of the solution is 8.30°
dlinn [17]

Answer: The mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=8.30^0C = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

K_f = freezing point constant =  

m= molality = \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}=\frac{74.8g\times 1000}{1450g\times 89.09g/mol}=0.579

8.30^0C=1\times K_f\times 0.579

K_f=14.3^0C/m

Let Mass of solute (KBr) = x g

8.3^0C=1.72\times 14.3\times \frac{xg\times 1000}{119g/mol\times 1450g}

x=58.2g

Thus the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

7 0
2 years ago
Lead(II) nitrate, Pb(NO3)2, and potassium iodide, KI<br><br><br> net ionic equation:
MissTica

Answer:

Pb²⁺ (aq) + 2I⁻ (aq) → PbI₂ (s)

General Formulas and Concepts:

  • Solubility Rules
  • Reaction Prediction

Explanation:

<u>Step 1: RxN</u>

Pb(NO₃)₂ (aq) + KI (aq) → PbI₂ (s) + KNO₃ (aq)

<u>Step 2: Balance RxN</u>

Pb(NO₃)₂ (aq) + 2KI (aq) → PbI₂ (s) + 2KNO₃ (aq)

<u>Step 3: Ionic Equations</u>

Total Ionic Equation:

Pb²⁺ (aq) + 2NO₃⁻ (aq) + 2K⁺ (aq) + 2I⁻ (aq) → PbI₂ (s) + 2K⁺ (aq) + 2NO₃⁻ (aq)

<em>Cancel out spectator ions.</em>

Net Ionic Equation:

Pb²⁺ (aq) + 2I⁻ (aq) → PbI₂ (s)

7 0
3 years ago
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