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2 years ago

All of the following are forms of radiation except _______ rays

Chemistry

2 answers:

Marina CMI [18]2 years ago

8 0

All of the following are forms of radiation except delta rays.

Ganezh [65]2 years ago

3 0

Answer:

delta

Explanation:

Different forms of radiation includes alpha particle, beta particle and gamma radiation.

Alpha particles is a helium nucleus with 2 protons and 2 neutrons. It carries a +2 charge.

Beta particles have the charge and mass of an electron. Beta particles are either negatively charged electrons or positively charged positrons.

Gamma are high energy electromagnetic radiation.

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When one atom loses an electron and another atom accepts that electron a(n) bond between the two atoms results?

morpeh [17]

When an element losses its electron its called a cation. When an element accepted that electron it called anion. This is called an ionic bond.

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2 years ago

What reaction would take place if a hot tungsten filament bulb was surrounded by air

expeople1 [14]

I don't think it wont be a big explosion

4 0

2 years ago

Read 2 more answers

5. Gabi has plans with her friends to go to a concert on her birthday in 4 days. She is so excited that she wants to know how ma

drek231 [11]

Answer:

So she is very anxious because she has to wait 345600 seconds

Explanation:

60 second = 1 minute

60 minute = 1 hour

1 hour has 3600 seconds (60*60)

24 hour = 1 day

3600 second * 24 hours =

1 day has 86400 seconds so in four days

86400 * 4 = 345600

7 0

2 years ago

A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at

Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0

2 years ago

When adjusted for any changes in ΔHΔH and ΔSΔS with temperature, the standard free energy change ΔG∘TΔGT∘Delta G_{T}^{\circ} at

STALIN [3.7K]

The equilibrium constant is 0.0022.

Explanation:

The values given in the problem is

ΔG° = 1.22 ×10⁵ J/mol

T = 2400 K.

R = 8.314 J mol⁻¹ K⁻¹

The Gibbs free energy should be minimum for a spontaneous reaction and equilibrium state of any reaction is spontaneous reaction. So on simplification, the thermodynamic properties of the equilibrium constant can be obtained as related to Gibbs free energy change at constant temperature.

The relation between Gibbs free energy change with equilibrium constant is ΔG° = -RT ln K

So, here K is the equilibrium constant. Now, substitute all the given values in the corresponding parameters of the above equation.

We get,

$1.22 * 10^{5} = - 8.314* 2400 * ln K$

$\\ 1.22 * 10^{5} = -19953.6 * ln K$

$ln K = \frac{-1.22*10^{5} }{19953.6} =-6.114\\\\k =e^{-6.114}=0.0022$

So, the equilibrium constant is 0.0022.

4 0

2 years ago