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3 years ago

The diagram below shows a group of water molecules.

Chemistry

1 answer:

denis-greek [22]3 years ago

7 0

Answer:

hydrogen bonds, the positive and negative charges of the hydrogen and oxygen atoms that make up water molecules makes them attracted to one another.

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A sample of oxygen gas was collected via water displacement. since the oxygen was collected via water displacement, the sample i

mylen [45]

The sample of oxygen gas was collected through water displacement. So, the gas collected will be a mixture of oxygen and water vapor.

Given that the total pressure of the mixture of gases containing oxygen and water vapor = 749 Torr

Vapor pressure of pure water at $26.4^{0}C$ =25.81mmHg

= $25.81 mmHg*\frac{1 Torr}{1 mmHg} =25.81 Torr$

According to Dalton's law of partial pressures,

Total pressure = Partial pressure of Oxygen gas + Partial pressure of water

749 Torr = Partial pressure of Oxygen gas + 25.81 Torr

Partial pressure of Oxygen gas = 749 Torr - 25.81 Torr = 723.19 Torr

Therefore the partial pressure of Oxygen gas in the mixture collected will be 723.19 Torr

6 0

3 years ago

Does wind speed affect evaporation? How?

garik1379 [7]

Answer:

Wind moving over a water or land surface can also carry away water vapor, essentially drying the air, which leads to increased evaporation rates.

3 0

3 years ago

Read 2 more answers

What is the volume in liters of 5.25 moles of He gas?

serious [3.7K]

Answer:

$\boxed {\boxed {\sf 117.6 \ L \ He}}$

Explanation:

Regardless of the type of gas, 1 mole at standard temperature and pressure (STP) occupies a volume of 22.4 liters. In this case the gas is helium (He).

We can set up a ratio.

$\frac { 22.4 \ L \ He}{ 1 \ mol \ He}$

Multiply by the given number of moles.

$5.25 \ mol \ He *\frac { 22.4 \ L \ He}{ 1 \ mol \ He}$

The moles of helium will cancel.

$5.25 *\frac { 22.4 \ L \ He}{ 1 }$

$5.25 * { 22.4 \ L \ He}$

Multiply.

$117.6 \ L \ He$

5.25 moles of helium gas at STP is 117.6 liters of helium.

5 0

3 years ago

What is the difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm

Zanzabum

Answer:

The difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm is :

Difference in mass = 985.32 - 984.5 = 0.82 g

Explanation:

Part I :

$n =\frac{3.01\times 10^{24}}{6.022\times 10^{23}}$

n = 4.9983

n = 4.99 moles

(Note : You can also take n = 5 mole )

Molar mass of gold = 196.96 g/mole

This means, 1 mole of gold(Au) contain = 196.96 grams

So, 4.99 moles of gold contain = $5\times 196.96$ g

4.99 moles of gold contain = 984.8 g

Mass of ${3.01\times 10^{24}}$ atoms of gold = 984.5 g

Part II :

Density of Gold = $19.32 g/cm^{3}$

Volume of the cuboid = $length\times breadth\times height$

Volume of the gold bar = $6.00\times 4.25\times 2.00$

Volume of the gold bar = 51 $cm^{3}$

Using formula,

$Density = \frac{mass}{Volume}$

$Mass = Density\times Volume$

$Mass = 19.32 \times 51$

Mass = 985.32 g

So, A gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm has mass of 985.32 g

Difference in mass = 985.32 - 984.5 = 0.82 g

3 0

3 years ago

The reaction of an acid and a base will always produce what kind of salt?

Illusion [34]

Answer:

the answer of the question is c

7 0

2 years ago