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3 years ago

Why do things dissolve so well in water ?

Chemistry

1 answer:

Goshia [24]3 years ago

4 0

Answer:

Explanation:

Water is called the universal solvent. It is a polar molecule (105 degree angle between the H atoms) that gives it a + and a - side so to speak....which allows it to 'pull apart' substances....overcome their intra-molecular attractions to each other ...i.e. disssovle them

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Which would lower the reaction rate?

saveliy_v [14]

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Explanation: Decreasing the temperature of the system would lower the reaction rate.

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is carried out adiabatically in a constant-volume batch reactor. The rate law is Plot and analyze the conversion, temperature, a

Sphinxa [80]

Answer:

See explaination

Explanation:

The mole balance for a constant-volume batch reactor is given such as, For a first-order isothermal reaction, the time to reach a given conversion is the same for constant-pressure and constant-volume reactors. Also, the time is the same for a reaction of any order if there is no change in the number of moles.

Please kindly check attachment for the step by step solution of the given problem.

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3 years ago

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olganol [36]

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3 years ago

Read 2 more answers

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 $Ag_2S$