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3 years ago

How many atoms of hydrogen are in 0.500 mol of ch3oh molecules?

1 answer:

8 0

In 1 mol of CH3OH, you have 4 H-atoms (because 3 H-atoms are attached to the C-atom, and one H-atom in the OH group). That means in 0.500 mol of CH3OH, you have 2 H-atoms since it is halved. And then we have Avogadro's constant: 6.02 * 1023.

The question asks for how many hydrogen atoms there are in 0.500 mol CH3OH. Using the numbers that we have (Avogadro's constant and no. of H-atoms), the answer of the question will be something like:

H-atoms in CH3OH = 2 * 6.02 * 1023 = ~1.2 * 1024

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Mendeleev created this table as he noticed that a

Nimfa-mama [501]

Answer:

Mendeleev had left the noble gases out of his periodic table.

Explanation:

Mendeleev's periodic table is pictured in the image attached to the question.

Mendeleev's table obviously lacked the noble gases. The reason for this grave omission is simple; the noble gases were not known as at the time when he formulated his periodic table. There weren't any known elements whose properties were similar to the properties of the noble gases. This would have lead him to suspect their existence.

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3 years ago

What is the electronic configuration of the following atoms?

Nastasia [14]

Answer:The electron configuration of an atom shows the number of electrons in each sublevel in each energy level of the ground-state atom. To determine the electron configuration of a particular atom, start at the nucleus and add electrons one by one until the number of electrons equals the number of protons in the nucleus. Each added electron is assigned to the lowest-energy sublevel available. The first sublevel filled will be the 1s sublevel, then the 2s sublevel, the 2p sublevel, the 3s, 3p, 4s, 3d, and so on. This order is difficult to remember and often hard to determine from energy-level diagrams such as Figure 5.8

A more convenient way to remember the order is to use Figure 5.9. The principal energy levels are listed in columns, starting at the left with the 1s level. To use this figure, read along the diagonal lines in the direction of the arrow. The order is summarized under the diagram

7 0

3 years ago

Chamber 1 and Chamber 2 have equal volumes of 1.0L and are assumed to be rigid containers. The chambers are connected by a valve

vitfil [10]

1) At tne same temperature and with the same volume, initially the chamber 1 has the dobule of moles of gas than the chamber 2, so the pressure in the chamber 1 ( call it p1) is the double of the pressure of chamber 2 (p2)

=> p1 = 2 p2

Which is easy to demonstrate using ideal gas equation:

p1 = nRT/V = 2.0 mol * RT / 1 liter

p2 = nRT/V = 1.0 mol * RT / 1 liter

=> p1 / p2 = 2.0 / 1.0 = 2 => p1 = 2 * p2

2) Assuming that when the valve is opened there is not change in temperature, there will be 1.00 + 2.00 moles of gas in a volumen of 2 liters.

So, the pressure in both chambers (which form one same vessel) is:

p = nRT/V = 3.0 mol * RT / 2liter

which compared to the initial pressure in chamber 1, p1, is:

p / p1 = (3/2) / 2 = 3/4 => p = (3/4)p1

So, the answer is that the pressure in the chamber 1 decreases to 3/4 its original pressure.

You can also see how the pressure in chamber 2 changes:

p / p2 = (3/2) / 1 = 3/2, which means that the pressure in the chamber 2 decreases to 3/2 of its original pressure.

5 0

3 years ago

Explain in a short sentence how you can tell if a reaction is a double replacement reaction.

Monica [59]

Explanation:

In a double displacement reaction, there is an actual exchange of partners to form new compounds.

The reaction is given as shown below:

AB + CD → AD + CB

One of the following conditions serves as the driving force for a double replacement reaction:

Formation of an insoluble compound or precipitate
Formation of water or any other non-ionizing compound
Liberation of a gaseous product.

4 0

3 years ago

Determine the freezing point and boiling point of a solution that has 68.4 g of sucrose

Ymorist [56]

Answer:

Freezing T° of solution = - 3.72°C

Boiling T° of solution = 101.02°C

Explanation:

To solve this we apply colligative properties. Firstly, freezing point depression:

ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Kf = Cryoscopic constant, for water is 1.86 °C/m

m = molality (moles of solute in 1kg of solvent)

i = Ions dissolved in solution

Our solute is sucrose, an organic compound so no ions are defined. i = 1.

Let's determine the moles: 68.4 g . 1mol/ 342g = 0.2 moles

molality = 0.2 mol / 0.1kg of water = 2 m

We replace data: ΔT = 1.86°C/m . 2m . 1

Freezing T° of solution = - 3.72°C

Now, we apply elevation of boiling point: ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of pure solvent

Kf = Ebulloscopic constant, for water is 0.512 °C/m

We replace:

Boiling T° of solution - Boiling T° of pure solvent = 0.512 °C/m . 2 . 1

Boiling T° of solution = 0.512 °C/m . 2 . 1 + 100°C → 101.02°C

6 0

3 years ago