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Lina20 [59]
3 years ago
14

We are given that when θ = 38.7°, the angle of refraction in the liquid is θ = 26.3°. Thus, from Snell's law, the index of refra

ction of the liquid is
Chemistry
1 answer:
oee [108]3 years ago
5 0

Answer:1.4

Explanation:

Angle of incidence= 38°

Angle of refraction=26.3°

From Snell's law: n= sin i/sinr

sin i= 0.6157

sin r= 0.4430

n= 0.6157/0.4430=1.4

Note, n is dimensionless

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Answer:

E) The rate of the reaction is directly proportional to the concentration of the reactant.

Explanation:

Give the characteristic of a first order reaction having only one reactant.

A) The rate of the reaction is not proportional to the concentration of the reactant.

B) The rate of the reaction is proportional to the square of the concentration of the reactant.

C) The rate of the reaction is proportional to the square root of the concentration of the reactant.

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E) The rate of the reaction is directly proportional to the concentration of the reactant.

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A vessel of volume 22.4 dm3 contains 20 mol h2 and 1 mol n2 ad 273.15 k initially. All of the nitrogen reacted with sufficient h
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Nitrogen combine with hydrogen to produce ammonia \text{NH}_3 at a 1:3:2 ratio:

\text{N}_2 \; (g) + 3 \;  \text{H}_2 \; (g) \leftrightharpoons 2\; \text{NH}_3 \; (g)

Assuming that the reaction has indeed proceeded to completion- with all nitrogen used up as the question has indicated. 3 \; \text{mol} of hydrogen gas would have been consumed while 2 \; \text{mol} of ammonia would have been produced. The final mixture would therefore contain

  • 17 \; \text{mol} of \text{H}_2 \; (g) and
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  • \begin{array}{lll} P(\text{H}_2) &= & n \cdot R \cdot T / V \\ & = & (17) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=&  1.723 \times 10^{3} \; \text{kPa} \end{array}
  • \begin{array}{lll} P(\text{NH}_3) &= & n \cdot R \cdot T / V \\ & = & (2) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=&  2.037 \times 10^{2} \; \text{kPa} \end{array}
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Now, the fact that we obtained 2-methylhexane upon hydrogenation and the two compounds had different heats of hydrogenation means that the two compounds were geometric isomers. The original compounds must have been cis-2-methyl-3-hexene and trans-2-methyl-3-hexene.

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