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vesna_86 [32]
3 years ago
5

Which of these particles is thought to carry the strong force?

Chemistry
2 answers:
kondor19780726 [428]3 years ago
8 0
Gluon because it is carried by a slew of particles.
Sidana [21]3 years ago
7 0
It’s gluon. Can you make me brainiest pls
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1) Increasing temperature

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The volume of a Goodyear Blimp is 2.5x10^7 L, which is occupied by 1.7x10^6 mol of helium. What is the internal pressure of the
attashe74 [19]

Answer:

1.7 bar

Explanation:

We can use the <em>Ideal Gas Law</em> to calculate the individual gas pressure.

pV = nRT     Divide both sides by V

 p = (nRT)/V

Data:   n = 1.7 × 10⁶ mol

R = 0.083 14 bar·L·K⁻¹mol⁻¹

T = 22 °C

V = 2.5 × 10⁷ L

Calculations:

(a) <em>Change the temperature to kelvins </em>

T = (22 + 273.15) K

  = 295.15 K

(b) Calculate the pressure

p = (1.7 × 10⁶ × 0.083 14 × 295.15)/(2.5× 10⁷)

  = 1.7 bar

4 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
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