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3 years ago

Determine the redox reaction represented by the following cell notation: Mg(s) | Mg2 (aq) || Cu2 (aq) | Cu(s)

1 answer:

6 0

Oxidation: Mg (s) —> Mg2+ (aq) + 2e
reduction: Cu2+ (aq) +2e —> Cu(s)

redox reaction: Mg(s) + Cu2+(aq)—> Mg2+(aq) + Cu(s)

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Why is U-235 the most common nuclear fuel instead of the more abundant U-238? State a clear topic sentence and support it with e

GREYUIT [131]

<h3>Answer and explanation;</h3>

The isotope U-235 is an important common nuclear fuel because under certain conditions it can readily be split, yielding a lot of energy. It is therefore said to be 'fissile' and use the expression 'nuclear fission'.
Uranium 238 on the other hand is not fissionable by thermal neutrons, but it can undergo fission from fast or high energy neutrons. Hence it is not fissile, but it is fissionable.
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3 years ago

H2SO4, sulfuric acid, contains three different types of atoms: hydrogen (H), sulfur (S), and oxygen (O). Each of these atoms rep

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Answer: Different elements. H2SO4 is a compound.

Explanation:

Element are the simplest substances and can't be broken down. Is a substance whose atoms all have the same number atomic number. The atomic number indicates the number of protons in the nucleus, and in a neutral atom it has the same number of neutrons. Therefore, for example, hydrogen is an element because it is an atom with a certain atomic number.

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4 years ago

How is the electron cloud model different from Bohr's atomic model?

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4 years ago

A 100mL reaction vessel initially contains 2.60x10^-2 moles of NO and 1.30x10^-2 moles of H2. At equilibrium the concentration o

Tcecarenko [31]

Answer:

$Kc=17.7$

Explanation:

Hello,

In this case, for the given reaction:

$2H_2(g) + 2NO(g)\rightleftharpoons 2 H_2O(g) + N_2(g)$

In such a way, the initial concentrations are:

$[H_2]_0=\frac{1.30x10^{-2}mol}{0.1L}=0.130M$

$[NO]_0=\frac{2.60x10^{-2}mol}{0.1L}=0.260M$

Thus, at equilibrium, the change $x$ , due to the chemical reaction extent, turns out:

$x=\frac{[NO]_{0}-[NO]_{eq}}{2} =\frac{0.260M-0.161M}{2}=0.049M$

Thus, the rest of the concentrations at equilibrium are:

$[H_2]_{eq}=0.130M-2(0.049M)=0.032M$

$[H_2O]_{eq}=2(0.049M)=0.098M$

$[N_2]_{eq}=0.049M$

In such a way, the equilibrium constant for the reaction, result as follows, even when on the statement the NO is excluded, because it participates in the equilibrium:

$Kc=\frac{[H_2O]_{eq}^2[N_2]_{eq}}{[H_2]_{eq}^2[NO]_{eq}^2}=\frac{(0.098M)^2(0.049M)}{(0.032M)^2(0.161M)^2} \\\\Kc=17.7$

Best regards.

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3 years ago

How to find protons neutrons and electrons of carbon?

Kryger [21]

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3 years ago