3 years ago

What is the lowest possible energy level that an electron can occupy?

1 answer:

6 0

Answer: the amount of energy needed to move an electron from one zone to another is a fixed, finite amount. At the lowest energy level, the one closest to the atomic center, there is a single 1s orbital that can hold 2 electrons

Explanation:

You might be interested in

The color of a star could give hints to mass temperature brightness density

Elza [17]

It’s basically just mass

7 0

3 years ago

Read 2 more answers

What is the energy of an electromagnetic wave with a electromagnetic wave with a frequency of 3 * 10 Hz ?

Olin [163]

Answer:

$E=1.989*10^{-32}J$

Explanation:

From the question we are told that:

Frequency $f=3 * 10 Hz$

Generally the equation for Energy of an electromagnetic wave is mathematically given by

$E=hf$

Where

h=Planck's constant

$h=6.62607015 * 10^{-34}$

Therefore

$E=6.62607015 * 10^{-34}*3 * 10$

$E=1.989*10^{-32}J$

8 0

3 years ago

Which of the following would be considered a solution?

sineoko [7]

The answer is sugar in water because it is soluble in water

4 0

4 years ago

Please help I’ll give you guys brainiest! Thank youuuuuuu :))

Nana76 [90]

Answer:

Balanced.

Explanation:

Both sides are pulling with equal force.

4 0

3 years ago

Read 2 more answers

What volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the su

anastassius [24]

Answer:

$V=27992L=28.00m^3$

Explanation:

Hello,

In this case, the combustion of methane is shown below:

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

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

$n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4$

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

$PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3$

Best regards.

8 0

4 years ago