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DochEvi [55]
2 years ago
9

Co (5. 00 g) and co2 (5. 00 g) were placed in a 750. 0 ml container at 50. 0 °c. The partial pressure of co in the container was

________ atm.
Physics
1 answer:
konstantin123 [22]2 years ago
6 0
Use PV = nRT and solve for P.
n = grams/molar mass = 5.00/molar mass CO2
Ignore the SO2.

But I think the answer it 6.74 atm

But I’m not saying but is but I’m just thinking this is the answer.

Goodluck!
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A puddle of water has a thin film of gasoline floating on it. A beam of light is shining perpendicular on the film. If the wavel
erastovalidia [21]

Answer:

D.

Explanation:

To solve the problem it is necessary to apply the concepts of Destructive and constructive interference. The constructive interference in tin film is given by

2t = (m+\frac{1}{2})\frac{\lambda}{n}

Where,

t = thickness

\lambda=Wavelenght

m= is an integer

n= film/refractive index

We use this equaton because phase change is only present for gasoline air interface, but not at the gasoline-water interface. <em>The minimum t only would be when the value of m=0 then</em>

2nt = \frac{\lambda}{2}

t = \frac{560nm}{4*1-4}

t = 100nm

Therefore the correct answer is D. The minimum thickness of the film to see ab right reflection is 100nm

4 0
3 years ago
One of your classmates, in a fit of unrestrained ego, jumps onto a lab table:
Anettt [7]

The equilibrium condition allows finding the results for the forces of the system are

      a) The free body diagram is in the attachment

      b) The normal force is N = 737 N

      c) The mass of the table is  10.2 kg

Newton's second law indicates that the net force is proportional to the product of the mass and the acceleration of the bodies, in the special case that the acceleration is zero, it is called the equilibrium condition

          ∑ F = 0

Where the bold letters indicate vectors, F is the external forces

a) A free body diagram is a scheme of the forces without the details of the bodies, in the attachmentt we see a free body diagram of the system.

b) The reaction force of the ground is applied in each of the legs of the table, in general this force has the same magnitude in each leg, therefore in Newton's second law we can place it as a single force

             N = N₁ + N₂ + N₃ + N₄₄

Let's apply the equilibrium condition

                N -  W_m -w_{table} = 0

                N =  W_m +w_{table}

                N = M_m g + w_{table}  

They indicate the pose of the boy is 65 kg, for the weight of the table of a laboratory table is approximately 100 N

                N = 65 9.8 + 100

                N = 737 N

c) To calculate the mass of the table we use the relation

                W = m_{table} g

                m_{table} = \frac{w_{table}}{g}

                m_{tabble}= \frac{100}{9.8}  

               m_{table}e = 10.2 kg

In conclusion using the equilibrium condition we can find the results for the forces are

      a) The free body diagram is in the attachment

      b) The normal force is N = 737 N

      c) The mass of the table is  10.2 kg

Learn more here:  brainly.com/question/19860811

7 0
2 years ago
If a particle with a charge of +4.3 × 10−18 C is attracted to another particle by a force of 6.5 × 10−8 N, what is the magnitude
AveGali [126]

Answer: 1.5×10^10 N/C

Explanation:

E= F/q

Where E= magnitude of the electric field

F= force of attraction

q= charge of the given body

Given F= 6.5×10^-8 N

q= 4.3× 10^-18 C

Therefore, E = 6.5×10 ^-8/ 4.3×10^-18

E = 1.5×10^10 N/C

7 0
3 years ago
The electric potential 1.34 m from a charge is 580 V. What is the value of the charge? Include the sign, + or -. (The answer is
blagie [28]

Answer: 8.6

Explanation:

5 0
2 years ago
for an ideal monoatomic gas, the internal energy U os due to the kinetic energy and U=3/2RT per mole.show that cv=3/2R per mole
sladkih [1.3K]

Answer:

i. Cv =3R/2

ii. Cp = 5R/2

Explanation:

i. Cv = Molar heat capacity at constant volume

Since the internal energy of the ideal monoatomic gas is U = 3/2RT and Cv = dU/dT

Differentiating U with respect to T, we have

= d(3/2RT)/dT

= 3R/2

ii. Cp - Molar heat capacity at constant pressure

Cp = Cv + R

substituting Cv into the equation, we have

Cp = 3R/2 + R

taking L.C.M

Cp = (3R + 2R)/2

Cp = 5R/2

3 0
2 years ago
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