How many moles of PCl3 are produced if 5 moles of P4 reacted with 22 moles of Cl2

Calculate the wavelength for the transition from n = 4 to n = 2, and state the name given to the spectroscopic series to which t

finlep [7]

Answer:

The wavelength for the transition from n = 4 to n = 2 is 486nm and the name name given to the spectroscopic series belongs to The Balmer series.

Explanation

lets calculate -

Rydberg equation- $\frac{1}{\pi } =R(\frac{1}{n_1^2} -\frac{1}{n_2^2})$

where , $\pi$ is wavelength , R is Rydberg constant ( $1.097\times10^7$ ), $n_1$ and $n_2$ are the quantum numbers of the energy levels. (where $n_1=2 , n_2=4$ )

Now putting the given values in the equation,

$\frac{1}{\pi }=1.097\times10^7\times(\frac{1}{2^2} -\frac{1}{4^2} )$ $=2056875m^-^1$

Wavelength $\pi =\frac{1}{2056875}$

= $4.86\times10^-^7$ = 486nm

 Therefore , the wavelength is 486nm and it belongs to The Balmer series.

8 0

3 years ago

AICI3(s)+Ca 3N2(s)-AIN(S)+CaCl2(s)

disa [49]

Answer:

2AlCl3 + Ca3N2 - 2AlN+ 3CaCl2

8 0

2 years ago

A chemist needs to make 250 mL of a 2.50 M aqueous solution of ammonium hydroxide from a 6.00 M ammonium hydroxide solution. How

Furkat [3]

Answer:

250 mL (total solution) = 104 mL (stock solution) + 146 mL (water)

Explanation:

Data Given

M1 = 6.00 M

M2 = 2.5 M

V1 = 250 mL

V2 = ?

Solution:

As the chemist needs to prepare 250 mL of solution from 6.00 M ammonium hydroxide solution to prepare a 2.50 M aqueous solution of ammonium hydroxide.

Now

first he have to determine the amount of ammonium hydroxide solution that will be taken from6.00 M ammonium hydroxide solution

For this Purpose we use the following formula

M1V1=M2V2

Put values from given data in the formula

6 x V1 = 2.5 x 250

Rearrange the equation

V1 = 2.5 x 250 /6

V1 = 104 mL

So 104 mL is the volume of the solution which we have to take from the 6.00 M ammonium hydroxide solution to prepare 2.5 M aqueous solution of ammonium hydroxide

But we have to prepare 250 mL of the solution.

so the chemist will take 104 mL from 6.00 M ammonium hydroxide solution and have to add 146 mL water to make 250 mL of new solution.

in this question you have to tell about the amount of water that is 146 mL

250 mL (total solution) = 104 mL (stock solution) + 146 mL (water)

7 0

3 years ago

True or False ... In a chemical change, a new substance is formed.

nikitadnepr [17]

The answer is true the answer has to be 20 words long so true true true true true true

7 0

3 years ago

How much thermal energy is added to 10.0 g of ice at −20.0°C to convert it to water vapor at 120.0°C?

Sonbull [250]

Answer:

7479 cal.

31262.2 joules

Explanation:

This is a calorimetry problem where water in its three states changes from ice to vapor.

We must use, the calorimetry formula and the formula for latent heat.

Q = m . C . ΔT

Q = Clat . m

First of all, let's determine the heat for ice, before it melts.

10 g . 0.5 cal/g°C ( 0° - (-20°C) = 100 cal

Now, the ice has melted.

Q = Clat heat of fusion . 10 g

Q = 79.7 cal/g . 10 g → 797 cal

We have water at 0°, so this water has to receive heat until it becomes vapor. Let's determine that heat.

Q = m . C . ΔT

Q = 10 g . 1 cal/g°C (100°C - 0°C) → 1000 cal

Water is ready now, to become vapor so let's determine the heat.

Q = Clat heat of vaporization . m

Q = 539.4 cal/g . 10 g → 5394 cal

Finally we have vapor water, so let's determine the heat gained when this vapor changes the T° from 100°C to 120°

Q = m . C . ΔT

Q = 10 g . 0.470 cal/g°C . (120°C - 100°C) → 94 cal

Now, we have to sum all the heat that was added in all the process.

100 cal + 797 cal + 1000 cal + 5394 cal + 94 cal =7479 cal.

We can convert this unit to joules, which is more acceptable for energy terms.

1 cal is 4.18 Joules.

Then, 7479 cal are (7479 . 4.18) = 31262.2 joules

6 0

3 years ago