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1 year ago

In the following equation, is being oxidized and is being reduced.

Chemistry

1 answer:

Mashcka [7]1 year ago

6 0

$oxidation \: number \: of \: oxygen = \\ before \: rxn = - 2 \\ after \: rxn = - 2$

$oxidation \: number \: of \: hydrogen = \\ before \: rxn = + 1 \\ after \: rxn = \\ 2x - 2 = 0 \\ x = + 1$

$oxidation \: number \: of \: carbon = \\ before \: rxn = \\ x - 6 = - 2 \\ x = 4 \\ after \: rxn = \\ x - 4 = 0 \\ x = 4$

<h2>Option A</h2>

$oxidation \: numbers \: remain \: constant \\ so \: none \:a re \: undergoing \: oxidation \: \\ nor \: reduction \:$

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5. The element copper has naturally occurring isotopes with mass numbers of 63

natali 33 [55]

Answer:

63.616

Explanation:

DATA

1. first atomic mass;m1=63

second atomic mass;m2=65
first percentage;p1= 69.2%
second percentage me;p2=30.8%
average mass;avg= ?

SOLUTION

avg= (m1)(p1) + (m2)(p2)

100

avg= (63)(69.2) + (65)(30.8)

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avg= 4359.6 + 2002

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4 0

2 years ago

Read 2 more answers

Calculate the energy required to ionize a ground state hydrogen atom. report your answer in kilojoules.

hjlf

First we find for the wavelength of the photon released due to change in energy level. We use the Rydberg equation:

1/ʎ = R [1/n1^2 – 1/n2^2]

where,

ʎ is the wavelength

R is the rydbergs constant = 1.097×10^7 m^-1

n1 is the 1st energy level = 1

n2 is the higher energy level = infinity, so 1/n2 = 0

Calculating for ʎ:

1/ʎ = 1.097×10^7 m^-1 * [1/1^2 – 0]

ʎ = 9.1158 x 10^-8 m

Then calculate the energy using Plancks equation:

E = hc/ʎ

where,

h is plancks constant = 6.626×10^−34 J s

c is speed of light = 3x10^8 m/s

E = (6.626×10^−34 J s * 3x10^8 m/s) / 9.1158 x 10^-8 m

E = 2.18 x 10^-18 J = 2.18 x 10^-21 kJ

This is still per atom, so multiply by Avogadros number = 6.022 x 10^23 atoms / mol:

E = (2.18 x 10^-21 kJ / atom) * (6.022 x 10^23 atoms / mol)

E = 1312 kJ/mol

3 0

3 years ago

The nucleus contains both protons and electrons True or false

Varvara68 [4.7K]

Answer: False

Explanation: The nucleus of an atom only contains the protons and neutrons.

The electrons are not found in the nucleus,

they are orbiting the nucleus in different shells.

3 0

3 years ago

What is easier to observe about an object,it’s physical properties or it’s chemical properties. Explain why.

Sladkaya [172]

Answer:

Physical Properties

Explanation:

You are able to see physical properties but are unable to see chemical properties.

7 0

2 years ago

The standard cell potential Ec for the reduction of silver ions with elemental copper is 0.46V at 25 degrees celsius. calculate

Cloud [144]

Answer : The $\Delta G$ for this reaction is, -88780 J/mole.

Solution :

The balanced cell reaction will be,

$Cu(s)+2Ag^+(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)$

Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half oxidation-reduction reaction will be :

Oxidation : $Cu\rightarrow Cu^{2+}+2e^-$

Reduction : $2Ag^++2e^-\rightarrow 2Ag$

Now we have to calculate the Gibbs free energy.

Formula used :

$\Delta G^o=-nFE^o$

where,

$\Delta G^o$ = Gibbs free energy = ?

n = number of electrons to balance the reaction = 2

F = Faraday constant = 96500 C/mole

$E^o$ = standard e.m.f of cell = 0.46 V

Now put all the given values in this formula, we get the Gibbs free energy.

$\Delta G^o=-(2\times 96500\times 0.46)=-88780J/mole$

Therefore, the $\Delta G$ for this reaction is, -88780 J/mole.

7 0

3 years ago

In the following equation, ______ is being oxidized and ______ is being reduced.

In the following equation, is being oxidized and is being reduced.