When the equation for the combustion of ethane, C2H6, is correctly balanced, the coefficient on oxygen is:

What is the associated deBroglie wavelength of a H2 molecule moving on one direction with kinetic energy of (3/2 kT) at 30 K

andrey2020 [161]

Answer: The de-Broglie's wavelength of a hydrogen molecule is $3.26\AA$

Explanation:

Kinetic energy is the measure of temperature of the system.

The equation used to calculate kinetic energy of a particle follows:

$E=\frac{3}{2}kT$

where,

E = kinetic energy of the particles = ?

k = Boltzmann constant = $1.38\times 10^{-23}J/K$

T = temperature of the particle = 30 K

Putting values in above equation, we get:

$E=\frac{3}{2}\times 1.38\times 10^{-23}J/K\times 30K\\\\E=6.21\times 10^{-22}J$

Calculating the mass of 1 molecule of hydrogen gas:

Conversion factor used: 1 kg = 1000 g

1 mole of hydrogen gas has a mass of 2 grams or $2\times 10^{-3}kg$

According to mole concept:

$6.022\times 10^{23}$ number of molecules occupy 1 mole of a gas.

As, $6.022\times 10^{23}$ number of hydrogen molecules has a mass of $2\times 10^{-3}kg$

So, 1 molecule of hydrogen will have a mass of = $\frac{2\times 10^{-3}kg}{6.022\times 10^{23}}\times 1=3.32\times 10^{-27}kg$

To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

$\lambda=\frac{h}{\sqrt{2mE_k}}$

where,

$\lambda$ = De-Broglie's wavelength = ?

h = Planck's constant = $6.624\times 10^{-34}Js$

m = mass of 1 hydrogen molecule = $3.32\times 10^{-27}kg$

$E_k$ = kinetic energy of the particle = $6.21\times 10^{-22}J$

Putting values in above equation, we get:

$\lambda=\frac{6.624\times 10^{-34}Js}{\sqrt{2\times 3.32\times 10^{-27}kg\times 6.21\times 10^{-22}J}}$

$\lambda=3.26\times 10^{-10}m=3.26\AA$ (Conversion factor: $1\AA=10^{-10}m$ )

Hence, the de-Broglie's wavelength of a hydrogen molecule is $3.26\AA$

3 0

3 years ago

Calculate the sublimation pressure of the solid at the melting point of 400.00 K assuming that the enthalpy of sublimation is no

nasty-shy [4]

This question is incomplete, the complete question is;

Tonksite is a solid at 300.00K. At 300.00 K its enthalpy of sublimation is 66.00 kJ/mol. The sublimation pressure at 300.00 K is 5.00 × 10⁻⁴ atm

Calculate the sublimation pressure of the solid at the melting point of 400.00 K assuming that the enthalpy of sublimation is not a function of temperature.

Answer: the sublimation pressure of the solid at the melting point is 0.3727 atm

Explanation:

Given that;

T1 = 300 K

T2 = 400 K

H_sub = 66 kJ/mol = 66000 J/mol

P1 = 5.00 × 10⁻⁴ atm

p2 = ?

now using the expression

log( p2 / 5.00 × 10⁻⁴ ) = (H_sub / R × 2.303 ) (( T2 - T1) / T1T2)

now we substitute of given values into the expression

log(p2/p1) = (66000 / 8.314 × 2.303 ) (( 400 - 300) / 300 × 400 )

p2 = 0.3727 atm

therefore the sublimation pressure of the solid at the melting point is 0.3727 atm

6 0

3 years ago

have you ever been swimming and noticed some areas in the water are colder and other are warmer? why do you think this happens

timama [110]

Maybe because someone hasn't been swimming on tht side bacause the water changes it's tempurture from your body heat or maybe because it isnt any sun out

6 0

3 years ago

Where does radioactivity come from?

Aleks04 [339]

Radioactivity comes from unstable atoms of certain elements. Radioactivity consists of alpha radiation (2 protons and 2 neutrons), beta radiation (1 electron), or gamma radiation (Electromagnetic photons).

8 0

3 years ago

Read 2 more answers

Hydrogen and oxygen react chemically to form water how much water would form if 14.8grams of hydrogen reacted with 34.8 grams of

pishuonlain [190]

Answer:

There will be formed 39.1935 grams H2O formed

Explanation:

Step 1: The balanced equation

2H2 + 02 → 2H20

Step 2: Given data

mass of hydrogen = 14.8 grams

Molar mass of hydrogen = 2.02 g/mole

mass of oxygen = 34.8 grams

Molar mass of oxygen = 32 g/mole

Step 3: Calculate moles

moles = mass / Molar mass

moles of hydrogen = 14.8g/ 2.02 g/mole = 7.33 moles

moles of oxygen = 34.8g / 32g/mole = 1.0875 moles

For 2 moles hydrogen consumed, we need 1 mole of oxygen.

This means oxygen is the limiting reagens and will be consumed completely. Hydrogen is the reactant in excess, there will remain 5.155 moles of hydrogen

Step 4: Calculate moles of H2O

We see that for 2 moles of H2 consumed, there is needed 1 mole of O2, to produce 2 moles of H2O.

For 1.0875 moles of oxygen consumed, there will be produced 2.175 moles of H2O

Step 5: Calculate mass of water

Mass of H2O = moles of H2O * Molar mass of H2O

Mass of H2O = 2.175 moles * 18.02 g/moles 39.1935 grams

There will be formed 39.1935 grams H2O formed

4 0

3 years ago