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3 years ago

Consider the following elementary reaction:

Chemistry

1 answer:

Fudgin [204]3 years ago

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Explanation:

The equation is given as;

N2O(g) ⇄ N2(g) + O(g)

k₁ = Forward reaction

k₋₁ = Reverse Reaction

Equilibrium concentration (K) = k₁ / k₋₁

$K = \frac{[N2O] }{[N2] [ O]}$

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The diameter of a biscuit is approximately 51 millimeters (mm). An atom of bismuth (Bi) is approximately 320. picometers (pm) in

astraxan [27]

Answer:

1.5e+8 atoms of Bismuth.

Explanation:

We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):

$\\ \frac{diameter\;biscuit}{diameter\;atom(Bi)}$

For this, it is necessary to know the values in meters for any of these diameters:

$\\ 1m = 10^{3}mm = 1e+3mm$

$\\ 1m = 10^{12}pm = 1e+12pm$

Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.

<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>

1 atom of Bismuth = 320pm in diameter.

$\\ 320pm*\frac{1m}{10^{12}pm} = 3.20*10^{-10}m$

<h3>Diameter of a biscuit in meters</h3>

$\\ 51mm*\frac{1}{10^{3}mm} = 51*10^{-3}m = 5.1*10^{-2}m$

<h3>Resulting Ratio</h3>

How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:

$\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1*10^{-2}m}{3.20*10^{-10}m}$

$\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}$

$\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{-2+10}$

$\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{8}=1.5e+8$

In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.

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3 years ago

Using the trend for chlorine, create a rule for naming halogen oxyacids.

Vlada [557]

All oxayacids have cations so no need to name the cation (H+) If name of polyatomic anion ends in -ate change to -ic for acid and if it ends with -ite change to -ous for acid

example:

ion nitrate is called nitric acid
ion nitrate is called nitrous acid

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3 years ago

Which atom would have the largest radius?

ryzh [129]

Answer:

D, Rubidium

Explanation

Rubidium- 290 pm

Cobalt- 200 pm

Beryllium- 153 pm

Helium- 140 pm

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