All categories

2 years ago

A 2.0 L container of nitrogen gas had a pressure of 3.2 atm. What volume would be necessary to decrease the pressure to 1.0 atm

Chemistry

1 answer:

beks73 [17]2 years ago

6 0

Answer:

6.4 L

Explanation:

When all other variables are held constant, you can use Boyle's Law to find the missing volume:

P₁V₁ = P₂V₂

In this equation, "P₁" and "V₁" represent the initial pressure and volume. "P₂" and "V₂" represent the final pressure and volume. You can find the theoretical volume by plugging the given values into the equation and simplifying.

P₁ = 3.2 atm P₂ = 1.0 atm

V₁ = 2.0 L V₂ = ? L

P₁V₁ = P₂V₂ <----- Boyle's Law

(3.2 atm)(2.0 L) = (1.0 atm)V₂ <----- Insert values

6.4 = (1.0 atm)V₂ <----- Simplify left side

6.4 = V₂ <----- Divide both sides by 1.0

You might be interested in

Find the volume of 56.0 grams of O2 at stp

BaLLatris [955]

The molar mass of O2 is 32g/mol. So the mol amount of these O2 is 56/32=2 mol. STP stands for the standard temperature and pressure which means the temperature is 0 ℃ and pressure is 100 kPa. And the molar volume of gas is 22.7 L/mol under STP. So the answer is 22.7*2=45.4 L

6 0

3 years ago

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]

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

<u>Explanation:</u>

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 ph of a buffer that is 0.145 m hc2h3o2 and 0.202 m kc2h3o2. The ka for hc2h3o2 is 1.8 × 10-5.

koban [17]

Answer:

pH = 4.8

Explanation:

A buffer is formed by a weak acid (0.145 M HC₂H₃O₂) and its conjugate base (0.202 M C₂H₃O₂⁻ coming from 0.202 M KC₂H₃O₂). The pH of a buffer system can be calculated using Henderson-Hasselbalch's equation.

$pH = pKa + log\frac{[base]}{[acid]} \\pH = -log(1.8 \times 10^{-5} )+log(\frac{0.202M}{0.145M} )\\pH=4.8$

6 0

2 years ago

The molar volume of oxygen,O2, is 3.90 dm3 mol-1 at 10.0 bar and 200 degree centigrade. Assuming that the expansion may be trunc

schepotkina [342]

Answer:

B = - 0.0326 dm³/mol

Explanation:

virial eq until second term:

PVm = RT [ 1 + B/Vm ]

∴ P = 10 bar * (atm/ 1.01325 bar) = 9.869 atm

∴ T = 200°C = 473 K

∴ Vm = 3.90 dm³/mol

∴ R = 0.08206 dm³.atm/K.mol

⇒ PVm / RT = 1 + B/Vm

⇒ ((9.869 atm)*(3.90 dm³/mol)) / ((0.08206 dm³.atm/mol.K)*(473K)) = 1 + B/Vm

⇒ 0.99164 = 1 + B/Vm

⇒ B/Vm = - 8.357 E-3

⇒ B = (3.90 dm³/mol)*( - 8.357 E-3 )

⇒ B = - 0.0326 dm³/mol

4 0

3 years ago

What is the bond order of li2−? express the bond order numerically?

jeka94

Atomic Number of Lithium is 3, so it has 3 electrons in its neutral state. Also, Li₂ will have 6 electrons. But the chemical formula we are given has a negative charge on it (i.e Li₂⁻) so there is an additional electron (RED) present on this compound. So, the total number of electrons are 7. The MOT diagram for this compound is shown below. According to diagram we are having 4 electrons in Bonding Molecular Orbitals (BMO) and 3 electrons in Anti-Bonding Molecular Orbitals (ABMO). Bond Order is calculated as,

Bond Order = (# of e⁻s in BMO - # of e⁻s in ABMO) ÷ 2

Bond Order = (4 - 3) ÷ 2

Bond Order = 1 ÷ 2
Or,
Bond Order = 1/2
Or,
Bond Order = 0.5

4 0

3 years ago