Helppp me plisss The chemistry

one kilogram of water (V1 = 1003 cm^3*kg-1) in a piston cylinder device at (25°C) and 1 bar is compressed in a mechanically reve

vova2212 [387]

Answer:

Q = -18118.5KJ

W = -18118.5KJ

∆U = 0

∆H = 0

∆S = -60.80KJ/KgK

Explanation:

W = RTln(P1/P2)

P1 = 1bar = 100KN/m^2, P2 = 1500bar = 1500×100 = 150000KN/m^2, T = 23°C = 23 + 273K = 298K

W = 8.314×298ln(100/150000) = 8.314×298×-7.313 = -18118.5KJ ( work is negative because the isothermal process involves compression)

∆U = Cv(T2 - T1)

For an isothermal process, temperature is constant, so T2 = T1

∆U = Cv(T1 - T1) = Cv × 0 = 0

Q = ∆U + W = 0 + (-18118.5) = 0 - 18118.5 = -18118.5KJ

∆H = Cp(T2 - T1)

T2 = T1

∆H = Cp(T1 - T1) = Cp × 0 = 0

∆S = Q/T

Mass of water = 1kg

Heat transferred (Q) per kilogram of water = -18118.5KJ/Kg

∆S = (-18118.5KJ/Kg)/298K = -60.80KJ/KgK

4 0

4 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

4 years ago

PLEASE HELP ME ITS DUE IN 1 HOUR PLZZZ

Ierofanga [76]

I can’t use evidence from the picture but the green slime from the pond that they are talking about is called Algae, which is alive and can grow quickly with sunlight or any other source of energy such as water and carbon dioxide. Hope this helps!!

5 0

3 years ago

Read 2 more answers

CH4(g)+O2(g)→CO2(g)+H2O(g). What mass of water is produced from the complete combustion of 5.00×10−3 g of methane?

bixtya [17]

Answer:

1gram of water

Explanation:

First balance the equation

Ch4+2O2>CO2+2H2O

Ratio is 1:2

500×10-3what about 2?

500×10-3×2=1g

1g of water

4 0

3 years ago

Will give brainliest, like, 5 stars and 5 points to best answer

Oliga [24]

evaporation systems allow for an endless source of water. you can grab cups of water straight from the sea or even a lake. the use of evaporation allows for you to drink water thats even healthier than getting it from a cloud and it will leave all of the bad parts that used to be in the water in the first container you pour into. this system is most useful in hot climates such as places near the equator.

6 0

3 years ago