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Vlada [557]
3 years ago
13

In a particular redox reaction, no2– is oxidized to no3– and cu2 is reduced to cu . Complete and balance the equation for this r

eaction in acidic solution. Phases are optional.
Chemistry
2 answers:
lianna [129]3 years ago
8 0

Reduction half reaction: Cu²⁺(aq) + 2e⁻ → Cu⁰(s).


Oxidation half reaction: NO₂⁻(aq) + H₂O(l) → NO₃⁻(aq) + 2H⁺(aq) + 2e⁻.  

Balanced chemical reaction:

Cu²⁺(aq) + NO₂⁻(aq) + H₂O(l) → Cu(s) + NO₃⁻(aq) + 2H⁺(aq).

Copper is reduced from oxidation number +2 (Cu²⁺) to oxidation number 0 (Cu) and nitrogen is oxidized from oxidation number +3 (in NO₂⁻) to oxidation number +5 (in NO₃⁻).


crimeas [40]3 years ago
8 0

Answer:

Cu²⁺(aq) + 3NO₂⁻(aq) ⇄ Cu⁰(aq) + 2NO₃⁻(aq)

Explanation:

A redox reaction is a reaction that occurs with oxidation and a reduction. The compound or the element that oxides are called reducted agent because it promotes the reduction of the other one. The compound or the element that reduces is called the oxidation agent, for the same reason.

In the oxidation, the substance increases its oxidation number: it loses electrons. The opposite occurs in reduction: there is a decrease in the oxidation number and the substance gain electrons.

So, the oxidation reaction is:

NO₂⁻(aq) ⇄ NO₃⁻(aq) + 2e⁻

And the reduction:

Cu²⁺(aq) + 2e⁻ ⇄ Cu⁰(aq)

Note that in the first one, the oxidation number of O is always -2, and the oxidation number of N was from +3 to +5.

The balanced reaction will be:

Cu²⁺(aq) + 3NO₂⁻(aq) ⇄ Cu⁰(aq) + 2NO₃⁻(aq)

The number of atoms of an element must be the same on each side of the equation.

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The size (radius) of an oxygen molecule is about 2.0 ×10−10m. Make a rough estimate of the pressure at which the finite volume o
belka [17]

Answer:

Explanation:

We can calculate the volume  of the oxygen molecule as the radius of oxygen molecule is given as 2×10⁻¹⁰m.

We know that volume=4/3×πr³

volume =4/3×π(2.0×10⁻¹⁰m)³

volume=33.40×10⁻³⁰m³

Volume of oxygen molecule=33.40×10⁻³⁰m³

we know the ideal gas equation as:

PV=nRT

k=R/Na

R=k×Na

PV=n×k×Na×T

n×Na=N

PV=Nkt

p is pressure of gas

v is volume  of gas

T is temperature of gas

N is numbetr of molecules

Na is avagadros number

k is boltzmann constant =1.38×10⁻²³J/K

R is real gas constant

So to calculate pressure using the  formula;

PV=NkT

P=NkT/V

Since there is only one molecule of oxygen so N=1

P=[1×1.38×10⁻²³J/K×300]/[33.40×10⁻³⁰m³

p=12.39×10⁷Pascal

8 0
3 years ago
Predict whether the changes in enthalpy, entropy, and free energy will be positive or negative for the boiling of water, and exp
Sedbober [7]

Answer:

ΔH > 0; ΔS >0; ΔG = 0

Not spontaneous when T < 100 °C;

         Equilibrium when T = 100 °C

      Spontaneous when T > 100 °C

Step-by-step explanation:

The process is

H₂O(ℓ) ⇌ H₂O(g)

ΔH > 0 (positive), because we must <em>add heat</em> to boil water

ΔS > 0 (positive), because changing from a liquid to a gas i<em>ncreases the disorder </em>

ΔG = 0, because the liquid-vapour equilibrium process is at <em>equilibrium</em> at 100 °C

ΔG = ΔH – TΔS

Both ΔH and ΔS are positive.

If T = 100 °C, ΔG =0. ΔH = TΔS, and the system is at equilibrium.


If T < 100 °C, the ΔH term will predominate, because T has decreased below the equilibrium value.

ΔG > 0. The process is not spontaneous below 100 °C.


If T > 100 °C, the TΔS term will predominate, because T has increased above the equilibrium value.

ΔG < 0. The process is spontaneous above 100 °C.

4 0
3 years ago
DDT:
Assoli18 [71]

Answer:

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8 0
3 years ago
Which of these solutions are basic at 25 °C? Solution A: [OH−]=3.13×10−7 M Solution C: [H3O+]=0.000747 M Solution B: [H3O+
solong [7]

Answer:

Are basic:

[OH⁻] = 3.13x10⁻⁷M and [H₃O⁺] = 9.55x10⁻⁹M

Explanation:

A solution is basic when pH = - log [H₃O⁺] is higher than 7.

It is possible to convert [OH⁻] to [H₃O⁺] using:

[H₃O⁺] = 1x10⁻¹⁴ / [OH⁻]

a. [OH⁻] = 3.13x10⁻⁷M

[H₃O⁺] = 1x10⁻¹⁴ / [3.13x10⁻⁷M]

[H₃O⁺] = 3.19x10⁻⁸M

pH = - log [H₃O⁺] = 7.50

[OH⁻] = 3.13x10⁻⁷M is basic

b. pH = -log [H₃O⁺] = - log 0.000747M = 3.13.

This solution is not basic

c. [H₃O⁺] = 9.55x10⁻⁹M

pH = 8.02

This solution is also basic.

8 0
3 years ago
How many grams of iki would it take to obtain a 100 ml solution of 0.300 m iki? how many grams of iki would it take to create a
GalinKa [24]

0.300 M IKI represents the concentration which is in molarity of a potassium iodide solution. This means that for every liter of solution there are 0.300 moles of potassium iodide. Knowing that molarity is a ratio of solute to solution.

By using a conversion factor:

100 ml x (1L / 1000 mL) x (0.300 mol Kl / 1 L) x (166.0g / 1 mol Kl) = 4.98 g

Therefore, in the first conversion by simply converting the unit of volume to liter, Molarity is in L where the volume is in liters. The next step is converted in moles from volume by using molarity as a conversion factor which is similar to how density can be used to convert between volume and mass. After converting to moles it is simply used as molar mass of Kl which is obtained from periodic table to convert from mole to grams.

In order to get the grams of IKI to create a 100 mL solution of 0.600 M IKI, use the same formula as above:

100 ml x (1L / 1000 mL) x (0.600 mol Kl / 1 L) x (166.0g / 1 mol Kl) = 9.96 g

3 0
3 years ago
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