All categories

2 years ago

When an electron in an atom spontaneously jumps from a higher energy state to a lower energy state, the atom?

1 answer:

6 0

If an atom suffers from a collision, that causes an electron to jump from a lower to higher state, it is called collisional excitation

You might be interested in

A sample of Hydrogen gas has a volume of 25.5 liters at a pressure of 15.0 kPa. If the temperature is kept constant and the pres

patriot [66]

Answer:

<h2>7.65 litres </h2>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

Since we're finding the new volume

$V_2 = \frac{P_1V_1}{P_2} \\$

15 kPa = 15,000 Pa

50 kPa = 50,000 Pa

We have

$V_2 = \frac{15000 \times 25.5}{50000} = \frac{382500}{50000} \\ = 7.65$

We have the final answer as

<h3>7.65 litres </h3>

Hope this helps you

5 0

2 years ago

The scientific method is sometimes seen as more of a cycle than a linear process. Which statement gives a major

irakobra [83]

Answer:

Explanation:

When a conclusions is shared it gives room for other scientist with other ideas or different perspective to raise up new questions

4 0

2 years ago

2. What is the electron configuration of an electrically neutral atom of magnesium?

alekssr [168]

Are u talking about electron sublevel config or where the electrons show in the "rings" of the atom

7 0

3 years ago

What is the excess reactant in the combustion of 23 g of methane in the open atmosphere?

Marta_Voda [28]

Answer : The excess reactant in the combustion of methane in opem atmosphere is $O_{2}$ molecule.

Solution : Given,

Mass of methane = 23 g

Molar mass of methane = 16.04 g/mole

The Net balanced chemical reaction for combustion of methane is,

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

First we have to calculate the moles of methane.

$\text{ Moles of methane}=\frac{\text{ Given mass of methane}}{\text{ Molar mass of methane}}$ = $\frac{23g}{16.04g/mole}$ = 1.434 moles

From the above chemical reaction, we conclude that

1 mole of methane react with the 2 moles of oxygen

and 1.434 moles of methane react to give $\frac{2moles\times 1.434moles}{1moles}$ moles of oxygen

The Moles of oxygen = 2.868 moles

Now we conclude that the moles of oxygen are more than the moles of methane.

Therefore, the excess reactant in the combustion of methane in open atmosphere is $O_{2}$ molecule.

6 0

3 years ago

A soccer player kicks a ball 6.5 meters. How much time is needed for the ball to travel this distance

Ahat [919]

It is 67 I think it is

8 0

2 years ago