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lianna
4 months ago
11

What is the symbol for the isotope of 58 co that possesses 33 neutrons?

Chemistry
2 answers:
KiRa [2.9K]4 months ago
6 0
The element with atomic number 58 is Cerium, meaning its symbol should be Ce rather than Co, which belongs to Cobalt with atomic number 27. Therefore, the notation for isotopes consists of the element's symbol accompanied by a superscript and a subscript, properly aligned. The superscript indicates the mass number.

Mass number = protons + neutrons = 58 + 33 = 91

The subscript denotes the atomic number, which is 58. This notation is illustrated in the attached image.

castortr0y [3K]4 months ago
3 0

The notation for the isotope:

\large {{{91} \atop {58}} \right Ce}

Additional clarification

Elements found in nature exist in several forms known as isotopes.

Isotopes are atoms distinguished by having the same atomic number but differing in neutron count.

This means that isotopes share the same Proton (Atomic Number).

Atomic mass refers to the mean mass of all isotopes.

To determine an atom's mass, the standard is set against 1 atom of carbon-12, which has a mass of 12 amu.

This results in atomic mass being expressed relative to the carbon-12 atom.

\large {\boxed {mass ~ average ~ atom ~ X ~ = ~ \frac {mass\:isotope ~ 1 + mass ~ isotope ~ 2} {number ~ whole ~ atom ~ X}}

An atomic mass unit (amu) represents a relative mass of 1/12 the weight of a carbon-12 atom.

\large {\boxed {Ar ~ X = \frac {mass ~ average ~ 1 ~ atom ~ X} {\frac {1} {12}. mass ~ 1 ~ atom ~ C-12}}}

The unit 'amu' has now transitioned to simply 'u'.

For instance, carbon possesses three isotopes: ₆¹²C12, ₆¹³C, and ₆¹⁴C.

The Atomic Number (Z) specifies the number of protons in an element's atom. When the atom carries no charge, its proton count matches its electron count, allowing the atomic number to also reflect the electron count.

Thus, atomic number = proton count = electron count.

Mass Number (A) is calculated by adding the protons and neutrons together.

Mass Number (A) = Proton count + Neutron count.

This leads to the relationship between atomic and mass numbers given by:

Atomic Number (Z) = Mass Number (A) - Neutron count.

In notation for the element:

\large {{{A} \atop {Z}} \right X}

Information:

X = symbol for the element

A = mass number

   = proton count + neutron count

Z = atomic number

   = proton count = electron count in neutral atoms

The symbol for the isotope with mass number 58 and 33 neutrons:

The number 58 identifies the Atomic Number (Z) = proton count

And calculating the mass number gives us:

A = proton count + neutron count

A = 58 + 33

A = 91

The element with atomic number 58 is undoubtedly Cerium (Ce).

<pTherefore, the notation for the isotope:

\large{\boxed{\bold{ {{{91} \atop {58}} \right Ce}}}}

Explore further

element  X

Keywords: the representation, the isotope, Ce, atomic number, mass number

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The standard cell potential (E°cell) for the reaction below is +0.63 V. The cell potential for this reaction is ________ V when
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Answer: The cell potential for the given reaction stands at 0.50 V

Explanation:

The provided cell reaction is:

Pb^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Pb(s)

The half-reactions are:

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Initially, we need to find the cell potential for this reaction.

Utilizing the Nernst equation:

E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}

where,

F = Faraday's constant = 96500 C

R = gas constant = 8.314 J/mol·K

T = room temperature = 25^oC=273+25=298K

n = electrons exchanged in oxidation-reduction = 2

E^o_{cell} = standard electrode potential for the cell = +0.63 V

E_{cell} = cell potential for the reaction =?

[Zn^{2+}] = concentration of Zn²⁺ = 3.5 M

[Pb^{2+}] = 2.0\times 10^{-4}M

Now substituting all known values into the equation, we arrive at:

E_{cell}=(+0.63)-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{3.5}{2.0\times 10^{-4}}

E_{cell}=0.50V

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alisha [2963]

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where

are the n-factor, molarity, and volume for the acid and

represent the n-factor, molarity, and volume for NaOH.

We know that:n_1,M_1\text{ and }V_1HCl

By substituting the known values into the equation, we get:

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