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V125BC [204]
3 years ago
13

Cesium metal is frequently used in photoelectric cells because the amount of energy necessary to eject electrons from a cesium s

urface is relatively small-only 206.5 kJ/mol. Part A What wavelength of light (in nanometers) does this correspond to?

Chemistry
1 answer:
dybincka [34]3 years ago
8 0

Answer:

Wavelength of light in (nm) = 579 nm

Explanation:

At first you find out the amount of energy needed to just eject one electron. This is given by \frac{Energy}{Avogadro's number}

this energy is given in question in kj/mole. This \frac{Energy}{Avogadro's number} is the work function of cesium for each electron is equal to the planc'k  einstein equation.

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4 years ago
What is the density of carbon dioxide gas at -25.2°C and 98.0 kPa? A. 0.232 g/L OB. 0.279 g/L OC. 0.994 g/L OD. 1.74 g/L O E. 2.
Black_prince [1.1K]

Answer:

E. 2.09\frac{g}{L}

Explanation:

From the ideal gasses equation we have:

PV=nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

The number of moles is also expressed as: n=\frac{mass}{Molar mass}

If replacing this in the ideal gasses equation we have:

PV=\frac{mass}{Molarmass}.RT

If we pass V to divide, we have:

P=\frac{mass}{V}.\frac{RT}{Molarmass}

And the density d = \frac{mass}{V}, so replacing, we have:

P=\frac{dRT}{M}

Solving for d, we have:

d=\frac{P.M}{R.T}

Now we have to be sure that we have the correct units, so we need to convert the units for pressure and temperature:

-Convert P=98kPa to atm

98.0kPa*\frac{0.00986923atm}{1kPa}=0.97atm

-Convert T=-25.2°C to K

-25.2^{o}C+273.15=247.95K

Finally we can replace the values in the equation:

d=\frac{(0.97atm)*(44.01\frac{g}{mol})}{(0.082\frac{atm.L}{mol.K})*(247.15K)}

d=2.09\frac{g}{L}

8 0
3 years ago
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