Why is an excited atom more reactive than a neutral?

Cierta cantidad de hidrogeno esta encerrada en una camara de platino a volumen constante. Cuando la camara se sumerge en un baño

salantis [7]

Answer:

-245.7°C es la temperatura del gas bajo 100 torr

Explanation:

Para resolver esta pregunta debemos hacer uso de la ley de Boyle que establece que la presión de un gas es directamente proporcional a la temperatura de este cuando el volumen permanece constante. La ecuación es:

P1T2 = P2T1

<em>Donde P es presión y T temperatura absoluta del estado inicial, 1, y final, 2.</em>

<em />

Reemplazando:

P1 = 1000torr

T2 = ? -Incógnita-

P2 = 100torr

T1 = 273K -Temperatura del hielo fundido = 0°C = 273K

1000torrT2 = 100torr273

T2 = 27.3K

27.3K - 273 =

<h3>-245.7°C es la temperatura del gas bajo 100 torr</h3>

<em />

3 0

3 years ago

Each year over _______ new organic compounds are produced.

antoniya [11.8K]

Each year over 100,000 new organic compounds are produce

7 0

3 years ago

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How many atoms are there in 3.559*10^-6 mol of krypton?

algol [13]

Answer:

The answer is

<h2> $2.143 \times {10}^{18} \: \: atoms$ </h2>

Explanation:

To find the number of atoms given the number of moles we use the formula

N = n × L

where

N is the number of entities

n is the number of moles

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question

$n = 3.559 \times {10}^{ - 6} \: mol$

Substitute the values into the above formula and solve

That's

<h3> $N = 3.559 \times {10}^{ - 6} \times 6.02 \times {10}^{23} \\ = 2.1425 \times {10}^{18}$ </h3>

We have the final answer as

<h3> $N = 2.143 \times {10}^{18} \: \: atoms$ </h3>

Hope this helps you

3 0

3 years ago

A student must make a 3 m acid solution using a 5 m acid solution. which of these is the safest way to make the solution?

weqwewe [10]

The motivation to abstain from adding water to concentrated acids is that, with a few acids, amid weakening, a considerable measure of warmth is discharged, by adding the corrosive to the water, the generally extensive measure of water will retain the warmth. On the off chance that you added water to concentrated corrosive when you initially beginning pouring the water, it could get sufficiently hot for the little measure of water that was filled all of a sudden bubble and splatter corrosive on you. Concentrated sulfuric corrosive is most famous for doing this, not all acids get that hot on weakening, but rather in the event that you make a propensity for continually adding the corrosive to water for every one of them, you can't turn out badly.

8 0

3 years ago

In this experiment, 0.070 g of caffeine is dissolved in 4.0 ml of water. The caffeine is then extracted from the aqueous solutio

Damm [24]

2.0ml of methylene chloride solution is used each time to extract caffeine from the aqueous solution.

Consider the concentration of caffeine obtained during each individual extraction from the aqueous solution to be C.

The total amount of caffeine obtained during each extraction is calculated as

(Total volume of water used to make up the caffeine aqueous solution) x (concentration of caffeine obtained during each individual extraction from the aqueous solution) + (Volume of methylene chloride added during each extraction x distribution coefficient of caffeine x concentration of caffeine obtained during each individual extraction from the aqueous solution)

Substituting these values we get

The total amount of caffeine obtained during each extraction

= (4.0×C )+ (2.0×4.6 × C)

= 13.2 C

The amount of caffeine remaining in the aqueous solution is calculated as

(Total volume of water used to make up the caffeine aqueous solution) x (concentration of caffeine obtained during each individual extraction from the aqueous solution)

Substituting these values we get

The amount of caffeine remaining in the aqueous solution = 4 × C

The fraction of caffeine remaining in aqueous solution is calculated as

= (The total amount of caffeine obtained during each extraction)/ (The amount of caffeine remaining in the aqueous solution)

=4.0 C/13.2 C

= 1/3.3.

Therefore the fraction of caffeine left in aqueous solution after 3 extractions is =(1/3.3)^3 =0.028

Therefore, the total amount of caffeine extracted

=0.070 × (1-(1/3.3)^3)

= 0.068 g

5 0

3 years ago

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