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Bond [772]
3 years ago
6

The distance between the image and the object in a plane mirror is 40cm if the distance of the object is reduced by 9.2cm, what

is the mirror?​
Chemistry
1 answer:
finlep [7]3 years ago
4 0

Answer:

<u></u>

  • <u>10.8 cm</u>

Explanation:

The question is:

<h2>What is the distance of the object from the mirror?</h2>

To answer this question, you must learn that, for plane mirrors, the distance of the object from the mirror is equal to the distance of the image formed behind the mirror.

For instance if you stand 1 meter away from a mirror (a plane one) the position of the image of you will be 1 meter behind the mirror.

Here, the initial distance between the image and the object is 40 cm. Then, the distance between the image and the miror is 20 cm, such as it is the distance between the object and the mirror.

That the distance of the object is reduced by 9.2 cm, must be understood as that the object was placed 9.2 cm closer to the mirror.

Then, the new distance of the object from the mirror is 20 cm - 9.2 cm = 10.8 cm. ← answer

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nalin [4]

Answer:

ΔH° = -11 kj

Explanation:

Step 1: Data given

1)   Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)    ΔH = −23 kJ

2)   3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g)   ΔH = −39 kJ

3)   Fe3O4(s) + CO(g) → 3 FeO(s) + CO2(g) ΔH = +18 kJ

Step 2: The balanced equation

FeO + CO → Fe + CO2

Step 3:  Calculate ΔH for the reaction FeO(s) + CO(g) → Fe(s) + CO2(g).

To get this equation, we need to combine the 3 equations

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<u>This equation we add to the second equation</u>

3Fe2O3(s) + CO(g) + 2Fe3O4(s) + 2CO(g) → 2Fe3O4(s) + CO2(g) + 6FeO(s) + 2CO2 (g)

3Fe2O3(s) + 3CO(g) →  3CO2(g) + 6FeO(s)

ΔH°  = 2*18 + (-39) = 36 - 39 =  -3 kJ

<u>This new equation we will divide by 3 </u>

3Fe2O3(s) + 3CO(g) →  3CO2(g) + 6FeO(s)

Fe2O3(s) + CO(g) →  CO2(g) + 2FeO(s)

ΔH°  =-3/3 = -1 kJ

<u>Now we will substract this new equation from the first equation</u>

Fe2O3(s) + 3CO(g) - Fe2O3(s) - CO(g) → 2Fe(s) + 3CO2(g) -2FeO(s) - CO2(g)

2CO(g)  + 2FeO(s)  → 2Fe(s) + 2CO2(g)

ΔH° = -23kJ +1kJ

ΔH° = -22 kj

<u>The next equation we will divide by 2</u>

2CO(g)  + 2FeO(s)  → 2Fe(s) + 2CO2(g)

CO(g)  + FeO(s)  → Fe(s) + CO2(g)

ΔH° = -22kJ /2

ΔH° = -11 kj

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