What is mixture of noodles in a bowl

How does energy flow through the reef ecosystem?

Free_Kalibri [48]

Photosynthesizing plants and algae convert light energy into chemical energy, which then gets passed through the food web to plant eaters, flesh eaters, and ultimately to scavengers and decomposers.

8 0

2 years ago

Which of the following leads to a higher rate of diffusion?

tiny-mole [99]

<h3><u>Answer;</u></h3>

Higher velocity of particles

<h3><u>Explanation;</u></h3>

The diffusion rate is determined by a variety of factors which includes;

Temperature such that the higher the temperature, the more kinetic energy the particles will have, so they will move and mix more quickly and the diffusion rate will be high.
Concentration gradient such that the greater the difference in concentration, the quicker the rate of diffusion.
Higher velocity of particles increases the diffusion rate as this means more kinetic energy by the particles and hence the particles will mix and move faster, thus higher diffusion rate.

5 0

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 $Ag_2S$