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3 years ago

How does pressure affect the solubility of gases?

1 answer:

3 0

Greater the pressure,greater the solubility of gas in a liquid.

Want an example?
Here it is, in the cold drinks we buy, CO2(carbon dioxide) is made soluble under high pressure in the liquid. As soon as we open these cans or bottles, the pressure releases.

Isn't it easy?

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CaCo3+2HCL Cacl2+h2o+co2

Rasek [7]

Answer:

Here

Explanation:

This is an acid-base reaction (neutralization): CaCO3 is a base, HCl is an acid.

6 0

2 years ago

MULTIPLE CHOICE

Flauer [41]

Answer:

its very simple ans we have 2 just multiply256

6 0

3 years ago

A sample of a compound that contains only the elements C, H, and N is completely burned in O2 to produce 44.0 g of CO2, 45.0 g o

PolarNik [594]

Explanation:

what you have to do is to multiply the given grams with the ratio of grams of that certain element in it's full compound in order to isolate the compound and get the element.

because the question did not specify how many grams of NO2 is formed, we can assume that the mass is neglegible, thus 1 mole was assigned to Nitrogen.

3 0

3 years ago

When a mixture of sulfur and metallic silver is heated, silver sulfide is produced. What mass of silver sulfide is produced from

IgorLugansk [536]

The question is incomplete, here is the complete question.

When a mixture of sulfur and metallic silver is heated, silver sulfide is produced. What mass of silver sulfide is produced from a mixture of 3.0g Ag and 3.0g $S_8$ .

Answer : The mass of silver sulfide is produced from a mixture is 3.44 grams.

Explanation : Given,

Mass of Ag = 3.0 g

Mass of $S_8$ = 3.0 g

Molar mass of Ag = 107.8 g/mole

Molar mass of $S_8$ = 256 g/mole

Molar mass of $Ag_2S$ = 247.8 g/mole

First we have to calculate the moles of Ag and $S_8$ .

$\text{ Moles of }Ag=\frac{\text{ Mass of }Ag}{\text{ Molar mass of }Ag}=\frac{3.0}{107.8g/mole}=0.0278moles$

$\text{ Moles of }S_8=\frac{\text{ Mass of }S_8}{\text{ Molar mass of }S_8}=\frac{3.0g}{256g/mole}=0.0117moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$16Ag(s)+S_8(s)\rightarrow 8Ag_2S(s)$

From the balanced reaction we conclude that

As, 16 mole of $Ag$ react with 1 mole of $S_8$

So, 0.0278 moles of $Ag$ react with $\frac{0.0278}{16}=0.00174$ moles of $S_8$

From this we conclude that, $S_8$ is an excess reagent because the given moles are greater than the required moles and $Ag$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Ag_2S$

From the reaction, we conclude that

As, 16 mole of $Ag$ react to give 8 mole of $Ag_2S$

So, 0.0278 moles of $Ag$ react to give $\frac{0.0278}{16}\times 8=0.0139$ moles of $Ag_2S$

Now we have to calculate the mass of $Ag_2S$

$\text{ Mass of }Ag_2S=\text{ Moles of }Ag_2S\times \text{ Molar mass of }Ag_2S$

$\text{ Mass of }Ag_2S=(0.0139moles)\times (247.8g/mole)=3.44g$

Therefore, the mass of silver sulfide is produced from a mixture is 3.44 grams.

4 0

3 years ago

Calculate the atomic mass of nitrogen if the two common isotopes of

Misha Larkins [42]

Answer:

Atomic mass of nitrogen = 14.0067 amu.

Explanation:

Isotopes can be defined as two or more forms of a chemical element that are made up of equal numbers of protons and electrons but different numbers of neutrons.

Generally, the isotopes of a chemical element have the same chemical properties because of their atomic number but different physical properties due to their atomic weight (mass number).

The two isotopes of nitrogen are nitrogen-14 and nitrogen-15.

Given the following data;

Relative abundance of N-14 = 99.63%

Atomic mass of N-14 = 14.003

Relative abundance of N-15 = 0.37%

Atomic mass of N-15 = 15.000

The atomic mass is;

14.003 × (99.63/100) + 15.000 × (0.37/100)

Atomic mass = 14.003 × (0.9963) + 15.000 × (0.0037)

Atomic mass = 13.9512 + 0.0555

Atomic mass = 14.0067 amu.

<em>Therefore, the atomic mass of nitrogen is 14.0067 amu. </em>

3 0

3 years ago