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3 years ago

12

When you heat a 50.0 L cylinder containing 10.0 moles of oxygen to 333 K, what is the pressure inside? (Use the ideal gas law PV

= nRT with R = 8.31 L-kPa/mol-K.)
166 kPa

553 kPa

12.5 kPa

5.46 kPa

1 answer:

hoa [83]3 years ago

8 0

Hope this helps you

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See if you can think of a way to prove that it is the oxygen in air, not nitrogen, that causes rusting.

allsm [11]

Answer:

rusting occur in oxygen and water coz when you put a nail into the water and you leave it three days you will the nail become brownish and that is due to oxygen and water

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A light source of wavelength, l, illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 1 eV. A second l

madreJ [45]

Explanation:

From first source, kinetic energy ( $K.E_{1}$ ) ejected is 1 eV and wavelength of light is $\lambda$ .

From second source, kinetic energy ( $K.E_{2}$ ) ejected is 4 eV and wavelength of light is $\frac{\lambda}{2}$ .

Relation between work function, wavelength, and kinetic energy is as follows.

K.E = $\frac{hc}{\lambda} - \phi$

where, h = Plank's constant = $6.63 \times 10^{-34}$ J.s

c = speed of light = $3 \times 10^{8}$ m/s

Also, it is known that 1 eV = $1.6 \times 10^{-19}$ J

Therefore, substituting the values in the above formula as follows.

From first source,

$K.E_{1}$ = $\frac{hc}{\lambda} - \phi$

1 eV = $1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\lambda} - \phi$

$1.6 \times 10^{-19} J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (1)

From second source,

$K.E_{2}$ = $\frac{hc}{\lambda} - \phi$

$4 \times 1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\frac{\lambda}{2}} - \phi$

$6.4 \times 10^{-19} J = \frac{2 \times 1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (2)

Now, divide equation (2) by 2. Therefore, it will become

${6.4 \times 10^{-19}J}{2} = \frac{2 \times 1.98 \times 10^{-25} J.m}{2\lambda} - \frac{\phi}{2}$

$3.2 \times 10^{-19}J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \frac{\phi}{2}$ ......... (3)

Now, subtract equation (3) from equation (1), we get the following.

$1.6 \times 10^{-19} = \frac{\phi}{2}$

$\phi$ = $3.2 \times 10^{-19}$

= 2 eV

Thus, we can conclude that work function of the metal is 2 eV.

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4 years ago

Find the relationships between the variables affected by gas:

butalik [34]

PV=nRT

Where:

P= pressure

V= volume

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To get number of particles : apply the following equation.

n = N/L

Where ;

n =mol

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4 years ago

Water beads up on waxy surfaces because of a degree of adhesion with the surface. True or False

Soloha48 [4]

False. Water is basically better at sticking to itself than it is in sticking to the wax.

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Approximately how much of the world’s oil and natural gas reserves are believed to be in the arctic? View Available Hint(s) Appr

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