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stiks02 [169]
3 years ago
5

Relation between glancing and angle of deviation​

Chemistry
1 answer:
shtirl [24]3 years ago
5 0

Answer:

the Glancing angle is the angle between the incident ray and plane mirror which is 90o in the given case. The angle between the direction of the incident ray and the reflected ray is the angle of deviation. Since the angle of deviation for a plane mirror is twice the glancing angle, the angle of deviation is 1800.

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A certain substance X condenses at a temperature of 123.3 degree C . But if a 650. g sample of X is prepared with 24.6 g of urea
Pavel [41]

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<u>The molal boiling point elevation constant is 1.59 ≈  1.6</u> Kkgmol^{-1}

Explanation:

To solve this question , we will make use of the equation ,

ΔT_{b} = i*K_{b} *m

<em>where , </em>

  • <em>ΔT_{b} is the change in boiling point of the substance X ( °C or K)</em>
  • <em>i is the Vant Hoff Factor which = 1 in this case ( no unit )</em>
  • <em>K_{b} is the mola boiling point elevation constant of X ( Kkgmol^{-1})</em>
  • <em>m is the molality of the solution which has (NH_{2})_{2} CO as the solute and  X as the solution (molkg^{-1})</em>

  1. ΔT_{b} = 124.3 -123.3 = 1 °C or K;
  2. i=1;
  3. m= \frac{moles of solute}{weight of solvent(kg)}molkg^{-1}

           ∴ m = \frac{\frac{24.6}{60} }{\frac{650}{1000} }

  • <em>as the weight of (NH_{2})_{2} CO is 60g and thus number of moles = \frac{24.6}{60}</em>
  • <em>and the weight of solvent in kg is \frac{650}{1000}</em>

    4. K_{b} ⇒ ?

∴

1=1*K_{b} *\frac{\frac{24.6}{60} }{\frac{650}{1000} }

⇒ K_{b} = 1.59 ≈ 1.6 Kkgmol^{-1}

4 0
3 years ago
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