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Vesnalui [34]
1 year ago
7

Please help/show work!

Chemistry
1 answer:
Keith_Richards [23]1 year ago
3 0

Answer:

7.00335g

Explanation:

n=\frac{V}{V_m} \\n = \frac{11.2}{22.4} \\n= 0.5 mol

n=\frac{m}{M} \\m=nM\\m=(0.5)(14.0067)\\m=7.00335g

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 <span>A solution is somthing desolved in somthing else. By desolved i mean it needs to have some particals ionized a solid you place in water that dissosiates (ions split apart from each other) makes a solution a good solution you can make in your kitchen is a salt-water solution, Put some regular table salt in a glass and stir it and you will notice the salt "disapears" what happens is the sodium ions and the chloride Ions seperate and 'hide' between water molocules. 

In basic terms only some substances can make a solutions others are refered to as insoluble as they can't be seperated in water or another solvent. In actuality however all ionic compounds (compounds that are composed of ions) are at least somewhat soluble, but don't dissociate well at all in some solvents. 

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5 0
2 years ago
What reaction type is depicted by the following equation? 2C2H2 + 5O2 4CO2 + 2H2O
Nonamiya [84]
The answer is combustion
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3 years ago
Nonmetallic elements form ions by _____ valence electrons to complete their outer shell.
Brums [2.3K]

<u>Answer:</u> Non-metals form ions when they gain electrons.

<u>Explanation:</u>

An ion is formed when an atom looses or gains electron.

  • When an atom looses electrons, it will form a positive ion known as cation.
  • When an atom gains electrons, it will form a negative ion known as anion.

Metals are the elements which have a tendency to loose electrons and thus they form cations. <u>Example:</u> Sodium will loose 1 electron to form Na^+ ion.

Non-metals are the elements which have a tendency to gain electrons and thus they form anions. <u>Example:</u> Fluorine will gain 1 electron to form F^- ion.

Hence, non-metals form ions when they gain electrons.

7 0
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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