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2 years ago

-273 degrees = absolute zero

Chemistry

2 answers:

lianna [129]2 years ago

6 0

Answer:

-273 degrees Celsius = zero degrees Kelvin = -459.67 Fahrenheit

Explanation:

lys-0071 [83]2 years ago

4 0

The answer is Kelvin

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What is the vapor pressure of an aqueous solution that has a solute mole fraction of 0.1000? The vapor pressure of water is 25.7

mafiozo [28]

Answer:

χsolvent = 1.0000 - 0.1000 = 0.9000

Use Raoult's Law:

Psolution = (χsolvent) (P°solvent)

x = (0.900) (25.756)

x = 23.18 mmHg (to four sig figs)

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3 years ago

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3 years ago

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Electrons involved in bonding between atoms are __________________

Vitek1552 [10]

The name given to these electrons are that they are valence electrons or binding electrons as these are directly involved in chemical Bonding and allow for different compounds to be made.

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3 years ago

What is the total probability of finding a particle in a one-dimensional box in level n = 4 between x = 0 and x = L/8?

Lubov Fominskaja [6]

Answer:

P = 1/8

Explanation:

The wave function of a particle in a one-dimensional box is given by:

$\psi = \sqrt \frac{2}{L} sin(\frac{n \pi x}{L})$

Hence, the probability of finding the particle in the one-dimensional box is:

$P = \int_{x_{1}}^{x_{2}} \psi^{2} dx$

$P = \int_{x_{1}}^{x_{2}} (\sqrt \frac{2}{L} sin(\frac{n \pi x}{L}))^{2} dx$

$P = \frac{2}{L} \int_{x_{1}}^{x_{2}} (sin^{2}(\frac{n \pi x}{L}) dx$

Evaluating the above integral from x₁ = 0 to x₂ = L/8 and solving it, we have:

$P = \frac{2}{L} [\frac{L}{16} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})]$

$P = \frac{1}{8} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})$

Solving for n=4:

$P = \frac{1}{8} (1 - 4\frac{sin(\frac{4 \pi}{4})}{4 \pi})$

$P = \frac{1}{8} (1 - \frac{sin (\pi)}{\pi})$

$P = \frac{1}{8}$

I hope it helps you!

7 0

2 years ago

4.41 g of propane gas (C3H8) is injected into a bomb calorimeter and ignited with excess oxygen, according to the reaction below

gayaneshka [121]

Answer : The heat of the reaction is -221.6 kJ

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter

$q_{rxn}=-q_{cal}$

$q_{cal}=c_{cal}\times \Delta T$

where,

$q_{rxn}$ = heat released by the reaction = ?

$q_{cal}$ = heat absorbed by the calorimeter

$c_{cal}$ = specific heat of calorimeter = $97.1kJ/^oC=97100J/^oC$

$\Delta T$ = change in temperature = $(T_{final}-T_{initial})=(27.282-25.000)=2.282^oC$

Now put all the given values in the above formula, we get:

$q_{cal}=(97100J/^oC)\times (2.282^oC)$

$q_{cal}=221582.2J=221.6kJ$

As, $q_{rxn}=-q_{cal}$

So, $q_{rxn}=-221.6kJ$

Thus, the heat of the reaction is -221.6 kJ

6 0

2 years ago