Why do lithium (Li) and sodium (Na) have similar properties?

78.6 grams of O2 and 67.3 grams of F2 are placed in a container with a volume of 40.6 L. Find the total pressure if the gasses a

saul85 [17]

1) List the known and unknown quantities.

Sample: O2.

Mass: 78.6 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

Sample: F2.

Mass: 67.3 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

2) Find the pressure of O2.

2.1- List the known and unknown quantities.

Sample: O2.

Mass: 78.6 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

2.2- Convert grams of O2 to moles of O2.

The molar mass of O2 is 31.9988 g/mol.

$mol\text{ }O_2=78.6\text{ }g*\frac{1\text{ }mol\text{ }O_2}{31.9988\text{ }g\text{ }O_2}=2.46\text{ }mol\text{ }O_2$

2.3- Set the equation.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

$PV=nRT$

2.4- Plug in the known quantities and solve for P.

$(P)(40.6\text{ }L)=(2.46\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)$

.

$P_{O_2}=\frac{(2.46\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)}{40.6\text{ }L}$ $P_{O_2}=1.57\text{ }atm$

The pressure of O2 is 1.57 atm.

3) Find the pressure of F2.

3.1- List the known and unknown quantities.

Sample: F2.

Mass: 67.3 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

3.2- Convert grams of F2 to moles of F2.

The mmolar mass of F2 is 37.9968 g/mol.

$mol\text{ }F_2=67.3\text{ }g\text{ }F_2*\frac{1\text{ }mol\text{ }F_2}{37.9968\text{ }g\text{ }F_2}=1.77\text{ }mol\text{ }F_2$

3.3- Set the equation.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

$PV=nRT$

3.4- Plug in the known quantities and solve for P.

$(P)(40.6\text{ }L)=(1.77\text{ }mol\text{ }F_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)$

.

$P_{F_2}=\frac{(1.77molF_2)(0.082057L*atm*K^{-1}*mol^{-1})(316.28K)}{40.6\text{ }L}$ $P_{F_2}=1.13\text{ }atm$

The pressure of F2 is 1.13 atm.

4) The total pressure.

Dalton's law - Partial pressure. This law states that the total pressure of a gas is equal to the sum of the individual partial pressures.

4.1- Set the equation.

$P_T=P_A+P_B$

4.2- Plug in the known quantities.

$P_T=1.57\text{ }atm+1.13\text{ }atm$ $P_T=2.7\text{ }atm$

The total pressure in the container is 2.7 atm.

5 0

1 year ago

The velocity of an electron that is emitted from a metallic surface by a photon is 3.6E3 km*s^-1. (a) What is the wavelength of

kaheart [24]

(a) The wavelength of the electron is 202.25885 nm

(b) The minimum energy required to remove the electron is 1.6565 × 10⁻¹⁷ J

(c) The wavelength of the causing radiation is approximately 8.84 nm

(d) X-ray

The question parameters are;

The given parameters of the electron are;

The velocity of the electron, v = 3.6 × 10³ km/s

(a) de Broglie wavelength is given as follows;

λ = h/(m·v)

Where;

λ = The wavelength of the wave

h = Planck's constant = 6.626 × 10⁻³⁴ J·s

m = The mass of the electron = 9.1 × 10⁻³¹ kg

Therefore, we get;

λ = 6.626 × 10⁻³⁴/(9.1 × 10⁻³¹ × 3.6 × 10⁶) = 202.25885 × 10⁻⁶

The wavelength, λ, of the electron is 202.25885 × 10⁻⁶ m = 202.25885 nm

(b) The energy required to remove the electron from the metal surface is known as the work function, W₀, which is given by the following formula

W₀ = h·f₀

Where;

f₀ = The threshold frequency

Given that the threshold frequency, f₀ = 2.50 × 10¹⁶ Hz, we have;

W₀ = 6.626 × 10⁻³⁴ J·s × 2.50 × 10¹⁶ Hz = 1.6565 × 10⁻¹⁷ J

The energy required to remove the electron from the metal surface, W₀ = 1.6565 × 10⁻¹⁷ J

(c) The wavelength of the radiation that caused the photoejection of the electron is given as follows;

The energy of the incoming photon, E = W₀ + (1/2)·m·v²

Where;

v = The velocity of the electron, and m = The mass of the electron

Therefore;

E = 1.6565 × 10⁻¹⁷ + (1/2) × 9.1 × 10⁻³¹ kg × (3.6 × 10⁶ m/s)² = 2.24618 × 10⁻¹⁷ J

We have;

E = h·f

∴ f = (2.24618 × 10⁻¹⁷ J)/(6.626 × 10⁻³⁴ J·s) = 3.38994869 × 10¹⁶ Hz

The speed of light, c = 299,792,458 m/s

From the equation for the speed of light, we have;

λ = c/f

∴ λ = (299,792,458 m/s)/(3.38994869 × 10¹⁶ Hz) = 8.84356919 nm ≈ 8.84 nm

The wavelength of the radiation that caused photoejection of the electron, λ $_{causing \ radiation}$ ≈ 8.84 nm

(d) The kind of electromagnetic radiation used which has a wavelength of 8.84 nm is the X-Ray which are electromagnetic radiation having wavelengths that extend from 10 picometers to 10 nanometers.

Learn more about De Broglie wavelength here;

brainly.com/question/19131384

6 0

3 years ago

What evidence is there that Potential energy is entering the system

Ulleksa [173]

Answer:

in particular 69-189 084 = 9283983

6 0

3 years ago

1. What is uncertainty in measurements?

Alex_Xolod [135]

Answer:

In metrology, measurement uncertainty is the expression of the statistical dispersion of the values attributed to a measured quantity.By international agreement, this uncertainty has a probabilistic basis and reflects incomplete knowledge of the quantity value. It is a non-negative parameter.

<h2>Hope it helps you.</h2>

8 0

3 years ago

B. What do you notice about the number of atoms in one mole?

Doss [256]

Answer:

one mole of atom of any element contains6.022×1033 atoms regardless of the type of elements the mass of one mole of an element depend on what that element is and is equal to atom mass of that element in gram

5 0

3 years ago