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k0ka [10]
3 years ago
15

Consider the equation below. Upper C a upper C upper O subscript 3 (s) double-headed arrow upper C a upper O (s) plus upper C up

per O subscript 2 (g). What is the equilibrium constant expression for the given reaction? K subscript e q equals StartFraction bracket upper C a upper O EndBracket over upper C a upper C upper O subscript 3 EndBracket EndFraction. K subscript e q equals StartFraction StartBracket upper C upper O subscript 2 EndBracket StartBracket upper C a upper O EndBracket over StartBracket upper C a upper C upper O subscript 3 EndBracket EndFraction. K subscript e q equals StartBracket upper C upper O subscript 2 EndBracket. K subscript e q equals StartFraction 1 over StartBracket upper C upper O subscript 2 EndBracket EndFraction.
Chemistry
2 answers:
stiks02 [169]3 years ago
8 0

Answer:K subscript e q equals StartFraction StartBracket upper C upper O subscript 2 EndBracket StartBracket upper C a upper O EndBracket over StartBracket upper C a upper C upper O subscript 3 EndBracket EndFraction

Explanation: the answer has it's root in Law of mass action which states that; the rate of a chemical reaction is directly proportional to the product of the concentrations of the reactants raised to their respective stoichiometric coefficients.

Len [333]3 years ago
8 0

Answer: is A

Explanation:pij

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Find the ratio v/cv/c for an electron in the first excited state (n = 2) of hydrogen.
dezoksy [38]

The answer is 0.365:100.

v/c ratio represents ratio of speed of an electron (v) to the speed of light (c).

How is the speed of an electron calculated?

  • The speed of an electron (v) is given by Bohr's model as-

v =\frac{1}{n}\; \frac{e^{2}}{4\pi \varepsilon _{0}}\times \frac{2\pi }{h}

Now, for the first excited state, n = 2.

e - Charge of electron = 1.6×10^{-19} C

h - Plank's constant = 6.6×10^{-34} J.s

ε₀- permittivity = 8.85×10^{-12}N^{-1}.C^{2}.m^{-2}

  • Put the above data in the formula-

v =\frac{1}{2}\; \frac{e^{2}}{4\pi \varepsilon _{0}}\times \frac{2\pi }{h}       =\frac{1}{4}\times \frac{(1.6\times 10^{-19})^{2}}{8.85\times 10^{-12}\times 6.6\times 10^{-34}}\\ \\     =0.01096\times 10^{8} \\      \\=1.096\times 10^{6}ms^{-1}

  • Now, the speed of light, c = 3.0×10^{8}\ m/s
  • Thus, the v/c ratio for an electron in the first excited state is calculated as-

\frac{v}{c} =\frac{1.096* 10^{6}\ m/s }{3.0 *10^{8}\ m/s }= \frac{0.365}{100}

  • Hence, the v/c ratio = 0.365:100.

To learn more about the speed of an electron (v), visit:

brainly.com/question/13198566

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6 0
1 year ago
List 3 ways the periodic table is useful<br> Help!
SashulF [63]

The periodic table arranges the elements into families and periods (vertical and horizontal rows). The elements in each family have similar properties. As you go across a row, the properties vary gradually from one element to the next. The table tells you what elements may have similar chemical and physical properties.The periodic table describes the atomic structure of all known elements. For instance, by looking at the periodic table, you can find out the atomic mass and the number of electrons the element has. Each element has its own separate set of such data. No two elements are the same.This is perhaps the most useful feature of the Periodic Table. It is an excellent reference tool. In one place, you can find many properties of an element.

6 0
3 years ago
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Please help I need a 100
svlad2 [7]

your answer would be C.

6 0
3 years ago
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The atomic nucleus contains?
ycow [4]
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8 0
3 years ago
A sample of Cd(OH)2 is added to pure water and allowed to come to equilibrium at 25ºC. The concentration of Cd +2 = 1.7 x 10 -5M
Archy [21]

Answer:

Ksp=2.0x10^{-14}

Explanation:

Hello there!

In this case, given the solubilization of cadmium (II) hydroxide:

Cd(OH)_2(s)\rightleftharpoons Cd^{2+}(aq)+2OH^-(aq)

The solubility product can be set up as follows:

Ksp=[Cd^{2+}][OH^-]^2

Now, since we know the concentration of cadmium (II) ions at equilibrium and the mole ratio of these ions to the hydroxide ions is 1:2, we infer that the concentration of the latter at equilibrium is 3.5x10⁻⁵ M. In such a way, the resulting Ksp turns out to be:

Ksp=(1.7x10^{-5})(3.4x10^{-5})^2\\\\Ksp=2.0x10^{-14}

Regards!

3 0
3 years ago
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