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3 years ago

Which chemical equation represents a redox reaction?

Chemistry

2 answers:

svet-max [94.6K]3 years ago

8 0

Hello. This question is incomplete. The full question is:

"Which chemical equation represents a redox reaction? A. Mg(ClO3)2 + 2HCl → MgCl2 + 2HClO3 B. CO + H2O → H2 + CO2 C. 2NH4NO3 + CuCl2 → 2NH4Cl + Cu(NO3)2 D. Na2SO3 + FeBr2 → 2NaBr + FeSO3 "

Answer:

B. CO + H2O → H2 + CO2

Explanation:

The reactions that involve loss and gain of electrons are called oxide-reduction reactions. In the oxidation reaction, electrons are lost, while the reduction reaction consists of gaining electrons.

Oxidation can occur in three circumstances: when oxygen is added to the substance, when a substance loses hydrogen or when the substance loses electrons. Example: fruit salads tend to darken when they come into contact with air, this is because oxygen acts to promote fruit oxidation. A tip to avoid this is to add lemon or orange juice, as the vitamin C present in citrus fruits prevents the oxidizing action of oxygen on the salad.

Reduction, in turn, is the reverse and also occurs in three ways: when a substance loses oxygen, when it gains hydrogen or when it gains electrons. Example: when the copper oxide (black) is placed in an appropriate apparatus (chamber) for its reduction to occur, the hydrogen gas comes in contact with the super heated copper oxide and, as a result, it loses oxygen and gradually goes away. turning pink as it is being reduced to copper.

An example of a redox reaction is CO + H2O → H2 + CO2

noname [10]3 years ago

5 0

Answer:

Assign oxidation numbers to all atoms in the equation.

Compare oxidation numbers from the reactant side to the product side of the equation.

The element oxidized is the one whose oxidation number increased.

Explanation:

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2 years ago

Read 2 more answers

Find the pH of a 0.010 M HNO2 solution.

lidiya [134]

Data:
Molar Mass of HNO2
H = 1*1 = 1 amu
N = 1*14 = 14 amu
O = 3*16 = 48 amu
------------------------
Molar Mass of HNO2 = 1 + 14 + 48 = 63 g/mol

M (molarity) = 0.010 M (Mol/L)

Now, since the Molarity and ionization constant has been supplied, we will find the degree of ionization, let us see:
M (molarity) = 0.010 M (Mol/L)
Use: Ka (ionization constant) = $5.0*10^{-4}$
$\alpha^2 (degree\:of\:ionization) = ?$

$Ka = M * \alpha^2$
$5.0*10^{-4} = 0.010* \alpha^2$
$0.010\alpha^2 = 5.0*10^{-4}$
$\alpha^2 = \frac{5.0*10^{-4}}{0.010}$
$\alpha^2\approx500*10^{-4}$

$\alpha\approx\sqrt{500*10^{-4}}$
$\alpha \approx 2.23*10^{-3}$

Now, we will calculate the amount of Hydronium [H3O+] in nitrous acid (HNO2), multiply the acid molarity by the degree of ionization, we will have:

$[ H_{3} O^+] = M* \alpha$
$[ H_{3} O^+] = 0.010* 2.23*10^{-3}$
$[ H_{3} O^+] \approx 0.0223*10^{-3}$
$[ H_{3} O^+] \approx 2.23*10^{-5} \:mol/L$

And finally, we will use the data found and put in the logarithmic equation of the PH, thus:

Data:
log10(2.23) ≈ 0.34
pH = ?
$[ H_{3} O^+] = 2.23*10^{-5}$

Formula:
$pH = - log[H_{3} O^+]$

Solving:
$pH = - log[H_{3} O^+]$
$pH = -log2.23*10^{-5}$
$pH = 5 - log2.23$
$pH = 5 - 0.34$
$\boxed{\boxed{pH = 4.66}}\end{array}}\qquad\quad\checkmark$

Note:. The pH <7, then we have an acidic solution.

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