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3 years ago

How much energy does an X-ray with an 8 nm (8 x 10-9 m) wavelength have?

Chemistry

1 answer:

Vlada [557]3 years ago

8 0

Answer: 2.48×10^-17 J

Explanation:

Given the following :

Wavelength = 8nm (8 x 10^-9 m)

Energy(e) of X-ray =?

Energy=[speed of light(c) × planck's constant (h)] ÷ wavelength

Speed of light = 3×10^8m/s

Planck's constant = 6.626×10^-34 Js

Wavelength = 8 x 10^-9 m

Energy = [(3×10^8) * (6.626×10^-34)] / 8 x 10^-9

Energy = [19.878×10^(8-34)] / 8 x 10^-9

Energy = 2.48475 × 10^(-26+9)

Energy = 2.48×10^-17 J

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storchak [24]

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2S + 3O 2 --->2SO 3 synthesis

Explanation:

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3 years ago

150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reac

Murljashka [212]

Answer:

$0.175\; \rm mol \cdot L^{-1}$ .

Explanation:

Magnesium chloride and silver nitrate reacts at a $2:1$ ratio:

$\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s)$ .

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

$\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}$ .

The precipitate silver chloride $\rm AgCl$ is insoluble in water and barely ionizes. Hence, $\rm AgCl\!$ isn't rewritten as ions.

Net ionic equation:

$\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}$ .

Calculate the initial quantity of nitrate ions in the mixture.

$\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}$ .

Since nitrate ions $\rm {NO_3}^{-}$ do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

$n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol$ .

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be $(1/2)$ of the concentration in the original solution.

$\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}$ .