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3 years ago

Which must be the same when comparing 1 mol of oxygen, O2, with 1 mol of carbon monoxide has, CO?

1 answer:

3 0

Answer: 1 mol of oxygen, O₂, and 1 mol of CO will have the same number of molecules, and the same number of atoms.

Justification:

Althought the question is too open, other answers may arise, the most remarkable similarity between the two compounds is that both are diatomic.

That means that both molecules oxygen, O₂, and carbon monoxide, CO have two atoms.

So, 1 mol of oxygen, O₂, and 1 mol of CO will have the same number of molecules, and the same number of atoms.

You must remember that 1 mol means a specific number. It is Avogadro's number, which is 6.022 × 10 ²³.

So 1 mol of CO and 1 mol of O₂ are the same number of representative particles: 6.022 ×10²³ molecules eac, and two times that number of atoms each (since each molecule has two atoms).

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51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b

Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

$a = \sqrt{8} r$

$a = \sqrt{8} \times 144 pm$

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}$

density d = 19.41 g/cm³

Similarly; For a body-centered cubic structure

$r = \dfrac{\sqrt{3}}{4}a$

where;

r = 144

$144 = \dfrac{\sqrt{3}}{4}a$

$a = \dfrac{144 \times 4}{\sqrt{3}}$

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V 3.68 × 10⁻²³ cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}$

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

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3 years ago

How would one convert 15 seconds to hours?

laiz [17]

Explanation:

a ...15 seconds×(3600 seconds hour)

8 0

2 years ago

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Name and describe three measures of central tendency used to summarize data.

Yuliya22 [10]

The answer is mean,mode and median

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3 years ago

What is the mass percent of potassium sulfate in solution if 78g of potassium sulfate is dissolved in 500 mL of water? (The dens

Debora [2.8K]

Answer:

13.5 %

Explanation:

First we calculate the mass of 500 mL of water, using its density:

Volume * Density = Mass

500 mL * 1.00 g/mL = 500 g

Then we calculate the mass percent of potassium sulfate, using the formula:

Mass of Potassium Sulfate / Total Mass * 100%

78 g / (78 + 500) g * 100 % = 13.5 %

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2 years ago

What are the products of the following combustion reaction? 2C2H2(g) + 5O2(g) →

kakasveta [241]

A combustion reaction of an will generally produce CO2 and H20 -- carbon dioxide and water and/or an oxide

looking at the combustion material C2H2, you know that the end products will be CO2 and H20, so the question is how much of each will you get

well, look at the total amount of carbon atoms, 2 C2, which means a total of 4 carbon atoms in this reaction, since only CO2 has carbon atoms, that means there must be 4 CO2 as an end product and 4 CO2 will use up 4 of 5 O2 molecule leaving only 1 O2 molecule for the H2 reaction.

now O2 has a total of 2 oxygen molecules whereas H20 has only a single oxygen molecule, hence the end product must have 2 H20

check that the H atoms balance out on both sides

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2 years ago

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