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3 years ago

How do the electron configurations within the same group of elements compare?

Chemistry

1 answer:

Dafna11 [192]3 years ago

3 0

The electronic configuration of the elements which share the group, must have the last electron in same block..

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A 100-watt light bulb radiates energy at a rate of 100 J/s. (The watt, a unit of power or energy over time, is defined as 1 J/s.

Semmy [17]

Answer

2.7956 * 10^19 photons

Givens

Wavelength = λ = 525 * 10^-9 meters [1 nmeter = 1*10^-9 meters]
c = 3 * 10^8 meters
E = ???
W = 100 watts
t = 1 second
h= plank's Constant = 6.26 * 10^-34 J*s

Formula

E = h * c / λ

W = E / t

Solution

E = 6.26 * 10^-34 j*s * 3 * 10^8 m/s /525 * 10^-9 (m)

The meters cancel out. So do the seconds. You are left with Joules as you should be.

E = 3.577 * 10^-18 Joules

What you have found is the energy of 1 photon.

Now you have to find the Joules from the watts.

W = E/t

100 * 1 second = 100 joules

1 photon contains 3.577 * 10 ^ - 18 Joules

x photon = 100 joules

1/x = 3.577 * 10^-18 / 100 Cross multiply

100 = 3.577 * 10 ^ - 18 * x Divide both sides by 3.577 * 10 ^ - 18

100/3.577 * 10 ^ - 18 = 3.577 * 10 ^ - 18x / 3.577 * 10 ^ - 18

2.7956 * 10^19 photons = x

7 0

3 years ago

A rigid tank that contains 2 kg of N2 at 25°C and 550 kPa is connected to another rigid tank that contains 4 kg of O2 at 25°C an

koban [17]

Answer:

The volume in the first tank = 0.32 $m^{3}$

The volume in the second tank = 2.066 $m^{3}$

The final pressure of the mixture = 203.64 K pa

Explanation:

<u>First Tank </u>

Mass = 2 kg

Pressure = 550 k pa

Temperature = 25 °c = 298 K

Gas constant for nitrogen = 0.297 $\frac{KJ}{Kg K}$

From the ideal gas equation

P V = m R T

550 × V = 2 × 0.297 × 298

V = 0.32 $m^{3}$

This is the volume in the first tank.

<u>Second tank</u>

Mass = 4 kg

Pressure = 150 K pa

Temperature = 25 °c = 298 K

Gas constant for oxygen = 0.26 $\frac{KJ}{Kg K}$

From the ideal gas equation

P V = m R T

150 × V = 4 × 0.26 × 298

V = 2.066 $m^{3}$

This is the volume in the second tank.

This is the iso thermal mixing. i.e.

$P_{3} V_{3} = P_{1} V_{1} + P_{2} V_{2}$ ----- (1)

$V_{3} = V_{1} + V_{2}$

$V_{3} = 0.32 + 2.066$

$V_{3} = 2.386 \ m^{3}$

Put this value in equation (1)

$P_{3}$ × 2.386 = 550 × 0.32 + 150 × 2.066

$P_{3}$ = 203.64 K pa

Therefore the final pressure of the mixture = 203.64 K pa

3 0

3 years ago

What do you call a circuit with only one path

DanielleElmas [232]

If I'm correct the answer should be a series circuit :) Hopefully this helps you out

3 0

3 years ago

Add a constant temperature when the volume of the gas is decreased what happens to its pressure

Margaret [11]

Answer:

<h2>Pressure will increase</h2>

Explanation:

At a constant temperature, the pressure of gas will increase proportional to the decrease in volume of the gas.

P1V1= P2V2

Decrease in volume result in increase in pressure as the equation has to hold true.

8 0

2 years ago

How many grams are in 1.6 moles of potassium bromide?

slava [35]

Answer:

190.4g

Explanation:

1.6mol of KBr (119.002g KBr/1 mol) = 190.4g

since you want to find grams, take the molar mass of KBr (119.002) per 1 mol and use it as your conversion factor (119.002g KBr/1 mol) which will then cancel out mols and leave you with grams.

6 0

3 years ago