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kobusy [5.1K]
3 years ago
14

Which electronic configuration matches the model?

Chemistry
1 answer:
antiseptic1488 [7]3 years ago
7 0

Answer:

option 2 is correct answer. its nitrogen.

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1 Point
inna [77]

Answer:

KINETIC ENERGY

Explanation:

3 0
3 years ago
A 4.215 g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of oxygen gas, producing 9.58
Scilla [17]

Answer:

\% O=27.6\%

Explanation:

Hello,

In this case, for the sample of the given compound, we can compute the moles of each atom (carbon, hydrogen and oxygen) that is present in the sample as shown below:

- Moles of carbon are contained in the 9.582 grams of carbon dioxide:

n_C=9.582gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2}  =0.218molC

- Moles of hydrogen are contained in the 3.922 grams of water:

n_H=3.922gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O} =0.436molH

- Mass of oxygen is computed by subtracting both the mass of carbon and hydrogen in carbon dioxide and water respectively from the initial sample:

m_O=4.215g-0.218molC*\frac{12gC}{1molC} -0.436molH*\frac{1gH}{1molH} =1.163gO

Finally, we compute the percent by mass of oxygen:

\% O=\frac{1.163g}{4.215g}*100\% \\\\\% O=27.6\%

Regards.

5 0
3 years ago
“The chemical formula in the structure of phosphorus is shown below. This represents a(n)....”
den301095 [7]

Answer:

element made up of four (4) similar atoms.

Explanation:

An element is made up of one kind of atom. A compound is made up of 2 or more different kinds of atoms that are chemically together.

5 0
3 years ago
In a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous methane and gaseous water in a 0.778 L
sergejj [24]

Answer:

the water concentration at equilibrium is

⇒ [ H2O(g) ] = 0.0510 mol/L

Explanation:

  • CH4(g) + H2O(g) ↔ CO(g) + 3H2(g)

∴ Kc = ( [ CO(g) ] * [ H2 ]³ ) / ( [ CH4(g) ] * [ H2O(g) ] ) = 0,30

  • equilibrium:

⇒ [ CO(g) ] = 0.206 mol / 0.778 L = 0.2648 mol/L

⇒ [ H2(g) ] = 0.187 mol / 0.778 L = 0.2404 mol/L

⇒ [ CH4(g) ] = 0.187 mol / 0.778 L = 0.2404 mol/L

replacing in Kc:

⇒ ((0.2648) * (0.2404)³) / ([ H2O(g) ] * 0.2404 ) = 0.30

⇒ 0.0721 [ H2O(g) ] = 3.679 E-3

⇒ [ H2O(g) ] = 0.0510 mol/L

7 0
3 years ago
Identify ways to reduce the risk of liquid bumping while heating.
postnew [5]

The most common way of preventing bumping is by adding one or two boiling chips to the reaction vessel. However, these alone may not prevent bumping and for this reason it is advisable to boil liquids in a boiling tube, a boiling flask, or an Erlenmeyer flask.

7 0
2 years ago
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