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ser-zykov [4K]
3 years ago
8

What element produces violet light of approximately 412 nm?

Chemistry
2 answers:
Natali [406]3 years ago
8 0
a. Hydrogen

Hydrogen has a line at 410 nm.
Mercury has a line at 405 nm.
Sodium and neon have no lines near 412 nm.
irinina [24]3 years ago
4 0

Answer:

a. Hydrogen

Explanation:

Emission spectrum of hydrogen produces four main visible wavelengths out of which one is violet light at 412 nm.

Mercury has visible prominent lines of blue, green and yellow-orange.

Sodium has two prominent very close yellow wavelengths in its emission spectrum.

Neon spectra also consists of green to red lines but not violet light.

Thus, correct option is a.

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Consider the following intermediate chemical equations.(IMAGE) -205.7 kJ -113.4 kJ -14.3 kJ 78.0 kJ
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Answer:

Approximately -205.7\; \rm kJ.

Explanation:

This question can be solved using Hess's Law.

Start by considering: how can the first three reactions (with known \Delta H values) be combined to produce the reaction \rm CH_4\; (g) + 4\; \rm Cl_2\; (g) \to CCl_4\; (g) + 4\; HCl\; (g)?

Here's one possible combination:

  • Include the first reaction once, without inverting.
  • Invert the second reaction and include it once.
  • Include the third reaction after multiplying all its coefficients by two.

In other words, if (1), (2), and (3) denote the three reactions with know \Delta H values, respectively, then 1 \times (1) - 1 \times (2) + 2\times (3) will give the required reaction \rm CH_4\; (g) + 4\; \rm Cl_2\; (g) \to CCl_4\; (g) + 4\; HCl\; (g).

By Hess's Law, the \Delta H value of the reaction \rm CH_4\; (g) + 4\; \rm Cl_2\; (g) \to CCl_4\; (g) + 4\; HCl\; (g) will thus be:

\begin{aligned}&1 \times \Delta H_1 - 1\times \Delta H_2 + 2\times \Delta H_3\\ &= 1 \times 74.6\; \rm kJ - 1 \times 95.7\; \rm kJ +2 \times (-92.3\; \rm kJ) \\ &= -205.7\; \rm kJ\end{aligned}.

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Answer:

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