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4 years ago

An oxidation-reduction reaction involves the

Chemistry

2 answers:

nadya68 [22]4 years ago

4 0

The best answer is (3)

In these kind of reactions, there is a transfer of electrons from one reactant to another. electrons are lost from one substance and gained by another.

Oxidation is loss of electrons from a substance, and Reduction is gain of electrons by a substance.

These two processes cannot occur without the other. If there is a reduction there must be an oxidation reaction and vice versa. The reactions usually occur simultaneously.

For example, table salt is formed by a redox reaction. Sodium is oxidized i.e. loses an electron (and becomes positively charged) while chlorine gas is reduced i.e. gains the electron (and become negatively charged). The result is formation of sodium chloride.

Arisa [49]4 years ago

4 0

An oxidation-reduction reaction (redox) is the transfer of electrons (it's an ionic bond). So that would be number 3.

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The symbol for the hydronium ion is _____.<br><br> H+<br> H 3 O-<br> H 3 O +<br> OH-

Brrunno [24]

The symbol for the hydronium ion is H3O+.
An hydronium ion is usually formed when an acid is present in water. The hydronium ion is made up of three atoms of hydrogen and one atom of oxygen, thus an hydronium ion is a water molecule which has gained an extra positive hydrogen ion.

4 0

3 years ago

Read 2 more answers

Write the expression for the equilibrium constant for this reaction. 2N205(g) <==> 4N02(g) + 02(g)

Elis [28]

Answer:

K = [NO₂]⁴[O₂] / [N₂O₅]²

Explanation:

Based on the chemical equilibrium reaction:

2 N₂O₅(g) ⇄ 4NO₂(g) + O₂(g)

The equilibrium reaction is obtained as the ratio between the multiplication of the concentrations of the reactants over the products powered to its reaction coefficient. That is:

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3 years ago

What is voltage? A. The pressure that pushes electrons to the anode; it is derived from the negative charge of electrons at the

Vlad [161]

Answer:

Explanation:

I think it's D but not 100% sure

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3 years ago

A solution is made containing 14.6g of CH3OH in 185g H2O.1. Calculate the mole fraction of CH3OH.2. Calculate the mass percent o

Andre45 [30]

Answer:

* $x_{CH_3OH}=0.0425$

* $\%m/m_{CH_3OH}=7.31\%$

* $m=2.46m$

Explanation:

Hello,

In this case, for the mole fraction of methanol we use the formula:

$x_{CH_3OH}=\frac{n_{CH_3OH}}{n_{CH_3OH}+n_{water}}$

Thus, we compute the moles of both water (molar mass 18 g/mol) and methanol (molar mass 32 g/mol):

$n_{CH_3OH}}=14.6g*\frac{1mol}{32g}=0.456molCH_3OH \\\\n_{water}}=185g*\frac{1mol}{18g}=10.3molH_2O$

Hence, mole fraction is:

$x_{CH_3OH}=\frac{0.456mol}{0.456mol+10.3mol}\\\\x_{CH_3OH}=0.0425$

Next, mass percent is:

$\%m/m_{CH_3OH}=\frac{m_{CH_3OH}}{m_{CH_3OH}+m_{water}}*100\%\\\\\%m/m_{CH_3OH}=\frac{14.6g}{14.6g+185g}*100\%\\\\\%m/m_{CH_3OH}=7.31\%$

And the molality, considering the mass of water in kg (0.185 kg):

$m=\frac{n_{CH_3OH}}{m_{water}} =\frac{0.456mol}{0.185kg}\\ \\m=2.46m$

Regards.

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3 years ago

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ASHA 777 [7]

A nuclear reaction in which a heavy nuclear splits spontaneously or on impact with another particle with the release of energy- fission

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