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2 years ago

What types of bonds or interactions are overcome when a nonmetal extended network melts?

Chemistry

1 answer:

STatiana [176]2 years ago

4 0

Covalent bonds or interactions are overcome when a nonmetal extended network melts.

Typically, nonmetals form covalent bonds with one another. A polyatomic ion's atoms are joined by a form of link called covalent bonding. A covalent bond requires two electrons, one from each of the two atoms that are connecting.

One technique to depict the formation of covalent connections between atoms is with Lewis dot formations. The number of unpaired electrons and the number of bonds that can be formed by each element are typically identical. Each element needs to share an unpaired electron in order to establish a covalent bond.

Therefore, covalent bonds or interactions are overcome when a nonmetal extended network melts.

Learn more about covalent bonds here;

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Copper is an example of a(n)

expeople1 [14]

copper is not a stone, copper is not a consumption resource, and copper is not renewable, that means that :

d. nonrenewable resource

Explanation:

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3 years ago

Read 2 more answers

Suppose a 250. mL flask is filled with 0.30 mol of N_2 and 0.70 mol of NO. The following reaction becomes possible:N_2(g) +O2 →

Inessa [10]

Answer:

0.4 M

Explanation:

Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.

Because there's no O₂ in the beginning, the NO will decompose:

N₂(g) + O₂(g) ⇄ 2NO(g)

0.30 0 0.70 Initial

+x +x -2x Reacts (the stoichiometry is 1:1:2)

0.30+x x 0.70-2x Equilibrium

The equilibrium concentrations are the number of moles divided by the volume (0.250 L):

[N₂] = (0.30 + x)/0.250

[O₂] = x/0.25

[NO] = (0.70 - 2x)/0.250

K = [NO]²/([N₂]*[O₂])

K = $\frac{(\frac{0.70 -2x}{0.250})^2 }{\frac{0.30+x}{0.250}*\frac{x}{0.250} }$

7.70 = (0.70-2x)²/[(0.30+x)*x]

7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)

4x² - 2.80x + 0.49 = 2.31x + 7.70x²

3.7x² + 5.11x - 0.49 = 0

Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70

x = 0.09 mol

Thus,

[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M

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3 years ago

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3 years ago

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3 years ago

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Hello!

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3 years ago