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timofeeve [1]
3 years ago
6

An object at a distance of 30 cm from a concave mirror gets its image at the same point. The focal length of the mirror is​

Chemistry
1 answer:
Nimfa-mama [501]3 years ago
3 0

The focal length of the mirror is 15 cm

From the question given above, the following data were obtained:

Object distance (u) = 30 cm

Image distance (v) = 30 cm

<h3>Focal length (f) =? </h3>

The focal length of the mirror can be obtained as follow:

\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\\\\frac{1}{f} = \frac{1}{30} + \frac{1}{30}\\\\\frac{1}{f} = \frac{2}{30} \\\\\frac{1}{f} = \frac{1}{15} \\\\

<h3>Invert </h3><h3>f = 15 cm</h3>

Therefore, the focal length of the mirror is 15 cm

Learn more: brainly.com/question/1392083

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<u>Answer:</u> The number of moles of HI in the solution is 1.24\times 10{-3} moles.

<u>Explanation:</u>

We are given:

K_c=7.00\times 10^{-5}\\n_{NH_3}=0.405mol\\n_{NH_4I}=1.45mol\\V=4.90L

To calculate the concentration of a substance, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     ......(1)

  • Concentration of ammonia:

[NH_3]=\frac{0.405mol}{4.90L}=0.083mol/L

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[NH_4I]=\frac{1.45mol}{4.90L}=0.30mol/L

For the given chemical reaction:

NH_4I(s)\rightleftharpoons NH_3(g)+HI(g)

The expression of K_c for above equation follows:

K_c=\frac{[HI][NH_3]}{[NH_4I]}

Putting values in above equation, we get:

7.0\times 10^{-5}=\frac{[HI]\times 0.083}{0.30}

[HI]=2.53\times 10^{-4}

Calculating the moles of hydrogen iodide by using equation 1, we get:

2.53\times 10^{-5}=\frac{\text{Moles of HI}}{4.9}\\\\\text{Moles of HI}=1.24\times 10^{-3}

Hence, the number of moles of HI in the solution is 1.24\times 10{-3} moles.

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