Reading glasses use what property of light waves to help people see better

Cual de los estados de agregacion es mas abundante en el universo

alisha [4.7K]

Answer: El Plasma, es el estado físico mas abundante no solo del universo sino de la naturaleza:

Explanation:

"El plasma es el estado de agregación más abundante de la naturaleza, y la mayor parte de la materia en el Universo visible se encuentra en estado de plasma, la mayoría del cual es el enrarecido plasma intergaláctico (particularmente el medio del intracluster) y en las estrellas."

7 0

3 years ago

What is the volume of an object with a mass of 31g and density of 444g/ml

Tanzania [10]

Answer:

<h2>The answer is 0.07 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

$volume = \frac{mass}{density} \\$

From the question

mass = 31 g

density = 444 g/mL

We have

$volume = \frac{31}{444} \\ = 0.06981...$

We have the final answer as

Hope this helps you

5 0

3 years ago

At elevations higher than sea level the atmospheric pressure is ___ than the atmospheric pressure at sea level

lions [1.4K]

Answer:

lower

Explanation:

The higher the sea level, the lower the atmospheric pressure. This is due to the density of air decreasing as the altitude increases.

5 0

3 years ago

If the temperature of a piece of steel decreases, what happens to its density? a. The density decreases. b. The density increase

vampirchik [111]

Answer:

Explanation:

4 0

3 years ago

Read 2 more answers

Gaseous methane (CH4) reacts with gaseous oxygen gas (02) to produce gaseous carbon dioxide (CO2) and gaseous water (H20). What

Mariana [72]

Answer:

Theoretical yield = 3.51 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For $CH_4$ :-

Mass of $CH_4$ = 1.28 g

Molar mass of $CH_4$ = 16.04 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{1.28\ g}{16.04\ g/mol}$

$Moles_{CH_4}= 0.0798\ mol$

For $O_2$ :-

Mass of $O_2$ = 10.1 g

Molar mass of $O_2$ = 31.998 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{10.1\ g}{31.998\ g/mol}$

$Moles_{O_2}= 0.3156\ mol$

According to the given reaction:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

1 mole of methane gas reacts with 2 moles of oxygen gas

0.0798 mole of methane gas reacts with 2*0.0798 moles of oxygen gas

Moles of oxygen gas = 0.1596 moles

Available moles of oxygen gas = 0.3156 moles

<u>Limiting reagent is the one which is present in small amount. Thus, $CH_4$ is limiting reagent. </u>

The formation of the product is governed by the limiting reagent. So,

1 mole of methane gas on reaction produces 1 mole of carbon dioxide.

0.0798 mole of methane gas on reaction produces 0.0798 mole of carbon dioxide.

Mole of carbon dioxide = 0.0798 mole

Molar mass of carbon dioxide = 44.01 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.0798\ moles= \frac{Mass}{44.01\ g/mol}$

Mass of $CO_2$ = 3.51 g

<u> Theoretical yield = 3.51 g</u>

3 0

3 years ago