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2 years ago

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which of the following properties of water is due primarily to the uneven distribution of charge between the hydrogen and oxygen

atoms? A. its color. B. its taste. C. its ability to dissolve ionic compounds. D. its ability to produce buoyant forces.

1 answer:

nydimaria [60]2 years ago

8 0

I am pretty sure that it's C.

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Serena mixed two samples together: a gray solid that smells like flowers and a gray solid that smells like bleach. She analyzed

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3 years ago

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2 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