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Alex17521 [72]
3 years ago
15

What do you do if your Bf call you selfish -_-

Chemistry
2 answers:
VLD [36.1K]3 years ago
7 0

Answer:

depends on what he called u selfish for

....

ryzh [129]3 years ago
4 0

Answer:

cry?!

Explanation:

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A sample of gas in a closed container at a temperature of 76°c and a pressure of 5.0 atm is heated to 399°c. What pressure does
NikAS [45]

<u>Answer:</u> The pressure that the gas exert at high temperature is 9.63 atm

<u>Explanation:</u>

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1\text{ and }T_1 are the initial pressure and temperature of the gas.

P_2\text{ and }T_2 are the final pressure and temperature of the gas.

We are given:

P_1=5.0atm\\T_1=76^oC=[76+273]K=349K\\P_2=?\\T_2=399^oC=[273+399]K=672K

Putting values in above equation, we get:

\frac{5.0atm}{349K}=\frac{P_2}{672K}\\\\P_2=9.63atm

Hence, the pressure that the gas exert at high temperature is 9.63 atm

6 0
3 years ago
Write the electron configuration for vanadium. (Use superscript and proper formatting for full credit.)
adoni [48]

Answer:

"Ar 3d3 4s2" is the electronic configuration for Vanadium.

8 0
3 years ago
Describe how Earth would be affected if it no longer got energy from the Sun.
ELEN [110]

Answer:

Fortunately, Earth retains heat fairly well, so humans wouldn't freeze instantly.  Without the Sun's rays, all photosynthesis on Earth would stop. All plants would die and, eventually, all animals that rely on plants for food — including humans — would die, too.

3 0
3 years ago
Read 2 more answers
How are physical and<br>chemical changes different?​
astra-53 [7]
Physical changes happen when you do/mix something that can be reversed. Best example is ripping a paper, you can just tape it back together so it is just an observable thing that changes, which is a physical change. Chemical changes happen when you mix things and they make a new substance and can’t be reversed. The best example of this is the Statue of Liberty, which changed in color due to the chemical change between the oxygen and copper, which created a new substance (rust) which made it green instead of copper.
Sorry for the long answer but hope this helps!
5 0
4 years ago
You calculate the Wilson equation parameters for the ethanol (1) 1 1-propanol (2) system at 258C and find they are L12 5 0.7 and
Gala2k [10]

Here is the correct question.

You calculate the Wilson equation parameters for the ethanol (1) +1 - propanol (2) system at 25° C   and find they are ∧₁₂ = 0.7 and ∧₂₁ = 1.1 . Estimate the value of parameters at 50° C

Answer:

the values of the parameters at 50° C are 0.766 and 1.047

Explanation:

From "critical point , enthalpy of phase change and liquid molar volume " the liquid molar volume (v) of ethanol and 1 - propanol is represented as follows:

Compound              Liquid molar volume    (cm³/mol)

Ethanol (1)                    58.68

1 - propanol (2)            75.14

To calculate the temperature dependent parameters of the Wilson equation ∧₁₂.

∧₁₂ = \frac{V_2}{V_1} \  exp \ (\frac{-a_{12}/R}{T} )          ------------ equation (1)

where:

a_{12}/R = Wilson parameter = ???

V_2 = liquid molar volume of component 2 = 75.14 cm³/mol

V_1 = liquid molar volume of component 1  = 58.68 cm³/mol

T = temperature  = 25° C  = ( 25 + 273.15) K = 298.15 K

Replacing our values in the above equation ; we have:

0.7 = \frac{75.14 \ cm^3/mol}{58.68 \ cm^3/mol} \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )

0.7 = 1.281 \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )

In (0.547) =  \ (\frac{-a_{12}/R}{298.15 \ K} )

-a_{12}/R=   0.60 * 298.15 \ K

-a_{12}/R=   - 178.89 \ K

a_{12}/R=    178.89 \ K

To calculate the temperature dependent parameters of the Wilson equation  ∧₂₁

∧₂₁ = \frac{V_1}{V_2} \  exp \ (\frac{-a_{12}/R}{T} )          ---------- equation (2)

1.1 = \frac{58.68 \ cm^3/mol}{75.14 \ cm^3/mol} \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )

1.1 = 0.7809 \ exp \  (\frac{-a_{12}/R}{298.15 \ K} )

\frac{1.1}{0.7809}=    exp \  (\frac{-a_{12}/R}{298.15 \ K} )

1n ( 1.4086)= (\frac{-a_{12}/R}{298.15 \ K} )

-a_{12}/R =     0.3426 * 298.15 \ K

-a_{12}/R =102.15 \ K

a_{12}/R = -102.15 \ K

From equation (1) ; let replace  178.98 K for a_{12}/R

V_2 = 75.14 cm³/mol

V_1 = 58.68 cm³/mol

T = 50° C = ( 50 + 273.15 ) K = 348.15 K

So;

∧₁₂ = \frac{75.14 \ cm^3/mol}{58.68 \ cm^3/mol} \ exp \ (\frac{- 178,.89 \ K}{348.15 \ K} )

∧₁₂ = 1.281 exp(-0.5138)

∧₁₂ = 1.281 × 0.5982

∧₁₂ =0.766

From equation 2; let replace 102.15 K for a_{12}/R

V_2 = 75.14 cm³/mol

V_1 = 58.68 cm³/mol

T = 50° C = ( 50 + 273.15 ) K = 348.15 K

So;

∧₂₁ = \frac{58.68 \ cm^3/mol}{75.14 \ cm^3/mol} \ exp \ (\frac{-(-102.15)\ K}{298.15 \ K} )

∧₂₁ =  0.7809 exp (0.2934)

∧₂₁ = 0.7809 × 1.3410

∧₂₁ = 1.047

Thus, the values of the parameters at 50° C are 0.766 and 1.047

8 0
3 years ago
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