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Kay [80]
3 years ago
9

Can someone shorten these definitions?

Chemistry
1 answer:
rodikova [14]3 years ago
5 0
A good night to see him again soon as I have to work tomorrow morning and I’m just gonna let go out and get it out and then I’ll head out and I can do that and then I’ll
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BRAINLIESTTT ASAP!! PLEASE HELP ME :)
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Answer: true

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A compound or material formed by a chemical reaction is known as the of that reaction.
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as the products of that reaction

Explanation:

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Use Lewis diagrams to show how electrons are shared to form covalent bonds in compounds wit the following atoms. Write the formu
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The Lewis structure/diagram for CH2O (aka Formaldehyde) can be written in either of the following ways shown in the picture.
The dots represent electrons in the valence shell of the atom (the outermost shell). The green dots are electrons that belong to the Oxygen atom, the blue belong to the Carbon atom, and the pink belong to the Hydrogen atoms.
Covalent bonds are bonds between atoms where atoms share electrons with each other. Atoms bond because they obey the octet rule ( the rule states that most atoms of main-group elements tend to want 8 electrons in their valence shells). 
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3 years ago
A 2.20 mol sample of NO 2 ( g ) is added to a 3.50 L vessel and heated to 500 K. N 2 O 4 ( g ) − ⇀ ↽ − 2 NO 2 ( g ) K c = 0.513
igor_vitrenko [27]

Answer:

[NO₂] = 0.434 M

[N₂O₄] = 0.0971 M

Explanation:

The equilibrum is:  N₂O₄(g)  ⇆  2NO₂ (g)

1 moles of nitrogen (IV) oxide is in equilibrium with 2 moles of nitrogen dioxide.

Initally we only have 2.20 moles of NO₂. So let's write the equilibrium again:

              2NO₂ (g)   ⇆   N₂O₄(g)      

Initially   2.20 mol              -

React          x                      x/2

X amount has reacted, and the half has been formed, according to stoichiometry.

Eq       (2.20-x) / 3.50L     (x/2)/ 3.50L

We divide by the volume because we need molar concentrations. Let's make the Kc's expression:

Kc = [N₂O₄] / [NO₂]²

0.513 = ((x/2)/ 3.50L) /  [(2.20-x) / 3.50L]

0.513 = ((x/2)/ 3.50L) / [(2.20-x)² / 3.50L²]

0.513 = ((x/2)/ 3.50L) / [2.20-x)² / 3.50L²]

0.513 = ((x/2)/ 3.50L) / (4.84 - 4.40x + x²) / 12.25)

0.513 / 12.25 (4.84 - 4.40x + x²) = x/2 / 3.50

0.203 - 0.184x + 0.0419x² = x/2 / 3.50

3.50(0.203 - 0.184x + 0.0419x²) = x/2

7 (0.203 - 0.184x + 0.0419x²) - x = 0

1.421 - 2.288x + 0.2933x² = 0  → Quadratic formula

a = 0.2933 ;  b = -2.288 ; c = 1.421

(-b +- √(b²-4ac)) / (2a)

x₁ = 7.12

x₂ = 0.68 → We consider this value, so we can have a (+) concentration.

Concentrations in the equilibrium are:

[NO₂] = (2.20-0.68) / 3.50 = 0.434 M

[N₂O₄] = (0.68/2) / 3.50  = 0.0971 M

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During the light-independent reaction, carbon dioxide is fixed by adding it to a A. 2-carbon compound B. 3-carbon compound C. 4-
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During the light independent reaction, carbon dioxide is fixed by adding it to a <span>5-carbon compound</span>
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