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Hunter-Best [27]

3 years ago

5

Visible light, radio waves, microwave radiation, infrared, ultraviolet radiation, X-rays, and gamma rays all

1 answer:

Softa [21]3 years ago

4 0

Answer:

they have equal energies

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A student finds that an unknown element reacts with elements from Group 2 (IIA). To which group does the unknown element most li

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Explanation:

A because a

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3 years ago

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Wind turbines are becoming an increasingly popular method of generating a community’s electricity. In wind turbines, wind is use

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The will not affect the wind. When it is windy the wind turns a turbine. Once the wind has gone through the turbine it is just normal wind, none of the wind is lost when this happened

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Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an

garri49 [273]

Explanation:

The given data is as follows.

T = $120^{o}C$ = (120 + 273.15)K = 393.15 K,

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So, $x_{1}$ = 0.5 and $x_{2}$ = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

$p^{sat}_{1}$ (393.15 K) = 9.2 bar

$p^{sat}_{1}$ (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

$p_{1} = x_{1} \times p^{sat}_{1}$

= $0.5 \times 9.2 bar$

= 4.6 bar

and, $p_{2} = x_{2} \times p^{sat}_{2}$

= $0.5 \times 10.5 bar$

= 5.25 bar

Therefore, the bubble pressure will be as follows.

P = $p_{1} + p_{2}$

= 4.6 bar + 5.25 bar

= 9.85 bar

Now, we will calculate the vapor composition as follows.

$y_{1} = \frac{p_{1}}{p}$

= $\frac{4.6}{9.85}$

= 0.467

and, $y_{2} = \frac{p_{2}}{p}$

= $\frac{5.25}{9.85}$

= 0.527

Calculate the dew point as follows.

$y_{1}$ = 0.5, $y_{2}$ = 0.5

$\frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}$

$\frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}$

$\frac{1}{P}$ = 0.101966 $bar^{-1}$

P = 9.807

Composition of the liquid phase is $x_{i}$ and its formula is as follows.

$x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{9.2}$

= 0.5329

$x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{10.5}$

= 0.467

4 0

4 years ago

Can someone help me with problem 21.105? Many thanks!

Reika [66]

The correct answer is 1 to the 3rd power

3 0

3 years ago

The mole fraction of a non-electrolyte (MM 40.0 g/mol) in a saturated aqueous solution is 0.310. What is the molality of the sol

jeka57 [31]

<u>Answer:</u> The molality of non-electrolyte is 24.69 m

<u>Explanation:</u>

We are given:

Mole fraction of saturated aqueous solution = 0.310

This means that 0.310 moles of non-electrolyte is present.

Moles of water (solvent) = 1 - 0.310 = 0.690 moles

To calculate the mass from given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of water = 0.690 moles

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

$0.690mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.690mol\times 18g/mol)=12.42g$

To calculate the molality of solution, we use the equation:

$\text{Molality}=\frac{n_{solute}\times 1000}{W_{solvent}\text{ (in grams)}}$

Where,

$n_{solute}$ = Moles of solute (non-electrolyte) = 0.310 moles

$W_{solvent}$ = Mass of solvent (water) = 12.42 g

Putting values in above equation, we get:

$\text{Molality of non-electrolyte}=\frac{0.310\times 1000}{12.42}\\\\\text{Molality of non-electrolyte}=24.96m$

Hence, the molality of non-electrolyte is 24.69 m

4 0

3 years ago