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Leni [432]
3 years ago
12

Fe₂O3 + co → Fe3O4 + CO₂

Physics
1 answer:
Goryan [66]3 years ago
6 0

Explanation:

                    Fe₂O₃  + CO  → Fe₃O₄ + CO₂

Balancing the equation above, we can derive simple mathematical equations that are very easy to solve.

             aFe₂O₃  + bCO  → cFe₃O₄ + dCO₂

a,b,c and d are the coefficients needed to balance the equation above;

  Conserving Fe; 2a = 3c

                       O: 3a + b = 4c + 2d

                        C: b = d

 let a = 1;

      c = \frac{2}{3}

      Since b = d

                  3a + d = 4c + 2d

                    3a = 4c + 2d - d

                     3a = 4c + d

           a = 1, c = \frac{2}{3}

                    3 = 4 x \frac{2}{3}  +  d

                   d = \frac{1}{3}

                    b = \frac{1}{3}

multiplying a, b, c and d by 3:

            a = 3    b = 1     c = 2   and d = 1

                  3Fe₂O₃  + CO  → 2Fe₃O₄ + CO₂

Learn more:

Balanced equation brainly.com/question/2612756

#learnwithBrainly

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How does paving parking lots and roads with concrete or asphalt affect surface water and groundwater?
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A car with a constant velocity of 22 m/s is driven for 6.8 s. How far did it travel? (v =∆ d/∆t)
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3 0
4 years ago
Part 1 :
dybincka [34]

Answer:

1. 218.55 N

2. 30.96^{o}

3. 2.1 m/s^{2}

Explanation:

Part 1;

Net force F=mg sin \theta where m is mass, g is gravitational force and \theta is the angle of inclination

F= 46*9.8*sin 29^{o}= 218.55N

Frictional force, F_{r} is given by

F_{r} = \mu_{s}mg cos \theta where \mu_{s} is the coefficient of static friction

F_{r} = 0.6*46*9.8 cos 29

F_{r}=236.57N

Since F_{r}>F, therefore, the block doesn’t slip and the frictional force acting is mgh=218.55N

Part 2.

Using the relationship that

Frictional force F_{s} = \mu_{s} mg cos \theta

mg sin \theta =\mu_{s} mg cos \theta

\mu_{s}= \frac {sin \theta}{cos \theta}

\mu_{s}= tan \theta

The maximum angle of inclination \theta = tan^{-1} \mu_{s}

\theta = tan^{-1} (0.6)

\theta= 30.96^{o}  

        

Part 3:

Net force on the object is given by

ma = mg sin 38 - \mu_{k} mg cos 38 where \mu_{k} is the coefficient of kinetic friction

 a = g ( sin 38 - \mu_{k} cos 38 )

                 = 9.8 ( sin 38 - (0.51) cos 38 )

                = 2.1m/s^{2}

7 0
4 years ago
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