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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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Calculate the energy (in kj/mol) required to remove the electron in the ground state for each of the following one-electron spec
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Explanation:

E_n=-13.6\times \frac{Z^2}{n^2}ev

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E_n = energy of n^{th} orbit

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a) Energy change due to transition from n = 1 to n = ∞ ,hydrogen atom .

Z = 1

Energy of n = 1 in an hydrogen like atom:

E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV

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E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0

Let energy change be E for 1 atom.

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Energy for 1 mole = E'

E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV

1 eV=1.60218\times 10^{-22} kJ

E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol

E'=1,312.17 kJ/mol

The energy  required to remove the electron in the ground state is 1,312.17 kJ/mol.

b) Energy change due to transition from n = 1 to n = ∞ ,B^{4+} atom .

Z = 5

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E_1=-13.6\times \frac{5^2}{1^2}eV=-340 eV

Energy of n = ∞ in an hydrogen like atom:

E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0

Let energy change be E.

E=E_{\infty}-E_1=0-(-340eV)=340 eV

1 mole = 6.022\times 10^{-23}

Energy for 1 mole = E'

E'=6.022\times 10^{-23} mol^{-1}\times 340eV

1 eV=1.60218\times 10^{-22} kJ

E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol

E'=32,804.31 kJ/mol

The energy  required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ ,Li^{2+}atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV

Energy of n = ∞ in an hydrogen like atom:

E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0

Let energy change be E.

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1 eV=1.60218\times 10^{-22} kJ

E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol

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The energy  required to remove the electron in the ground state is 11,809.55 kJ/mol.

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E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV

Energy of n = ∞ in an hydrogen like atom:

E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0

Let energy change be E.

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1 mole = 6.022\times 10^{-23}

Energy for 1 mole = E'

E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV

1 eV=1.60218\times 10^{-22} kJ

E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol

E'=820,107.88 kJ/mol

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We then use the conversion;

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