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Marianna [84]
3 years ago
5

For the reaction N2(g) + 3H2(g) <->2NH3(g), what will happen if hydrogen gas was removed from the mixture?

Chemistry
1 answer:
OlgaM077 [116]3 years ago
7 0
There will be a shift towards the reactants because the system will try to increase the amount of hydrogen. 
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What is the frequency of a photon with an energy of 3.26 x 10-19 J?
Kobotan [32]

Answer:

B.

from \: eisteins \: energy \: relation :  \\ E = hf  \\ h \: is \: plancks \: constant\\ 3.26 \times  {10}^{ - 19}  = 6.63 \times  {10}^{ - 34 }  \times f \\ f =  \frac{3.26 \times  {10}^{ - 19} }{6.63 \times  {10}^{ - 34} }  \\ f = 4.92 \times  {10}^{14}  \: Hz

8 0
2 years ago
Which of the following describes the relationship correctly?
galben [10]
A. Air expands, becomes less dense and rises when the temperature is high due to the increased kinetic energy. This causes the air molecules to move around much more so they are further apart.
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3 years ago
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HELPPPPPPPPPPPPPPPPPPPPPPPP MEEHHHHHHHHHHHHHH
AnnZ [28]
Here you are looking on the Free Body diagram of a net force of 0N in both the x and y-directions.  the only ones that has that condition met is A and C.
4 0
3 years ago
How many moles of FeCl3 could be produced from 6.1 moles of Cly?<br> 2 Fe + 3Cl2 —&gt; 2 FeCl2
Black_prince [1.1K]

Answer:

4.1 moles of FeCl₃

Explanation:

The reaction expression is given as shown below:

         2Fe    +       3Cl₂   →    2FeCl₃

Number of moles of Cl₂  = 6.1moles

So;

  We know that from the balanced reaction expression:

                       3 moles of Cl₂ will produce 2 moles of FeCl₃

Therefore    6.1moles of Cl₂ will produce \frac{6.1 x 2}{3}   = 4.1 moles of FeCl₃

The number of moles is 4.1 moles of FeCl₃

6 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
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