Ask question

Ask question

All categories

Murljashka [212]

3 years ago

7

How are the conditions at which phases are in equilibrium represented on a phase diagram?

1 answer:

insens350 [35]3 years ago

7 0

How are the conditions at which phases are in equilibrium represented on a phase diagram?

Image result for How are the conditions at which phases are in equilibrium represented on a phase diagram?

Along the line between liquid and solid, the melting temperatures for different pressures can be found. The junction of the three curves, called the triple point, represents the unique conditions under which all three phases exist in equilibrium together. Phase diagrams are specific for each substance and mixture.

You might be interested in

6. A fellow chemist in your research group made 2.4 moles of gas held at a

Whitepunk [10]

Ignore this its not the anwser 2.7g

7 0

3 years ago

Helppp ! Please :’)

Over [174]

Answer:

$\boxed{\sf \lambda \: = 1.282 \times {10}^{ 3} nm}$

Explanation:

Suppose an electron makes transition from n initial (ni) to n final (nf) then formula for wavelength is given by,

$\frac{1}{ \lambda} = R_H {Z}^{2} \big(\frac{1}{{n_f^2}} - \frac{1}{{n_i^2}} \big)$

Where,

λ is wavelength of photon
Rʜ is rydberg constant, the value of Rʜ is

109690 Cm-¹ in <em>Puri Sharma Pathania</em> standard book of physical chemistry &

109737 Cm-¹ according to <em>Wikipedia</em>

Z is the atomic number of atom, for hydrogen Z =1,

& according to given data, ni = 5, nf = 3

Solution:

Let's solve for wavelength,

Substituting all the given data in above formula,

$\frac{1}{ \lambda} = 109690 \times {1}^{2} \big(\frac{1}{{3^2}} - \frac{1}{{5^2}} \big)$

$\frac{1}{ \lambda} = 109690 \times {1}^{2} \big(\frac{1}{{9}} - \frac{1}{{25}} \big)$

$\frac{1}{ \lambda} = 109690 \times {1}^{2} \times \frac{25 - 9}{25 \times 9}$

$\frac{1}{ \lambda} = 109690 \times {1}^{2} \times \frac{16}{225}$

$\frac{1}{ \lambda} = 7800.18 \: \: cm ^{ - 1}$

$\lambda = 1.282 \times {10}^{ - 4} cm$

Now we know that, 1 cm = 10000000 nm,

Converting the wavelength from Cm → Nm

$\lambda \: = 1.282 \times {10}^{ - 4} \times {10}^{7}$

$\lambda \: = 1.282 \times {10}^{ - 4 + 7}$

$\sf \lambda \: = 1.282 \times {10}^{ 3} nm$

$\sf \small Thanks \: for \: joining \: brainly \: community!$

4 0

2 years ago

A student needs to make 0.855 L of a sodium bromide solution for an experiment. The concentration of the required solution is 0.

d1i1m1o1n [39]

Answer:

Explanation:

Molarity = moles/liters of solution

moles = molarity x Liters of solution

= 0.350 m/L x 0.855 L

= 0.299 mol

mass = moles x molar mass

= 0.299 mol x 102.89 g/ms

= 30.78 grams

4 0

3 years ago

What is the complete electron configuration of tin?

Leya [2.2K]

Kr 4d10 5s2 5p2 would be your answer.

5 0

4 years ago

The reaction of solid aluminum with hydrochloric acid is used to make hydrogen gas in a laboratory experiment. The reaction is 2

Gemiola [76]

Answer:

0.003088 moles of hydrogen gas were formed .

Explanation:

Pressure at which hydrogen gas is collected at 20°C = 768.0 Torr

Vapor pressure of water at 20°C = 17.5 Torr

Total pressure = Vapor pressure of water + Partial pressure of hydrogen gas

Partial pressure of hydrogen gas:

Total pressure - Vapor pressure of water

= 768.0 Torr - 17.5 Torr = 750.5 Torr = 0.987 atm

(1 Torr = 0.001315 atm)

Pressure of hydrogen gas =P = 0.986 atm

Temperature at which gas was collected ,T= 20°C = 293.15 K

Volume of the gas ,V= 75.3 mL = 0.0753 L

Moles of hydrogen gas = n

PV=nRT (An ideal gas equation)

$n=\frac{PV}{RT}=\frac{0.987 atm\times 0.0753 L}{0.0821 atm L/mol K\times 293.15 K}=0.003088 mol$

0.003088 moles of hydrogen gas were formed .

7 0

4 years ago