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3 years ago

What is the oxidation state of Sin SO 3 ^ 2- ?

Chemistry

1 answer:

Semmy [17]3 years ago

3 0

Answer:

Oxidation state of S in SO3^2- is +4

Explanation:

S in SO3^2-=

S = ?

O = -2

S + 3 moles of O * -2 = -2

S + (-6) = -2( it has charge -2)

S - 6 = -2

S = -2 + 6

S = +4

Note, the sign is very important

Don't confuse SO3 together with SO3^2-, the oxidation state of S in SO3 is + 6, how?

S = ?

O = -2

S + 3 * -2 = 0 ( because it has no charge)

S - 6 = 0

S = +6

And in SO3^2-

S = +4

The difference is that one has charge SO3^2- and the other doesn't SO3.

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3 years ago

The atomic symbol Superscript 206 subscript 82 upper P b. represents lead-206 (Pb-206), an isotope that has 82 protons and 124 n

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Answer:

\left \{ {{y=206} \atop {x=82}}Pb \right.

Explanation:

isotopes are various forms of same elements with different atomic number but different mass number.

Radioactivity is the emission of rays or particles from an atom to produce a new nuclei. There are various forms of radioactive emissions which are

Alpha particle emission \left \{ {{y=4} \atop {x=2}}He \right.
Beta particle emission \left \{ {{y=0} \atop {x=-1}}e \right.
gamma radiation \left \{ {{y=0} \atop {x=0}}γ \right.

in the problem the product formed after radiation was Pb-206. isotopes of lead include Pb-204, Pb-206, Pb-207, Pb-208. they all have atomic number 82. which means the radiation cannot be ∝ or β since both radiations will alter the atomic number of the parent nucleus.

Only gamma radiation with \left \{ {{y=0} \atop {x=0}}γ \right. will produce a Pb-206 of atomic number 82 and mass number 206 , since gamma ray have 0 mass and has 0 atomic number.equation is shown below

\left \{ {{y=206} \atop {x=82}}Pb\right ⇒ \left \{ {{y=206} \atop {x=82}}Pb\right + \left \{ {{y=0} \atop {x=0}}γ\right.

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4 years ago

Using the Bohr model, determine the energy, in joules, necessary to ionize a ground-state hydrogen atom. Show your calculations.

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Answer:

The energy required to ionize the ground-state hydrogen atom is 2.18 x 10^-18 J or 13.6 eV.

Explanation:

To find the energy required to ionize ground-state hydrogen atom first we calculate the wavelength of photon required for this operation.

It is given by Bohr's Theory as:

1/λ = Rh (1/n1² - 1/n2²)

where,

λ = wavelength of photon

n1 = initial state = 1 (ground-state of hydrogen)

n2 = final state = ∞ (since, electron goes far away from atom after ionization)

Rh = Rhydberg's Constant = 1.097 x 10^7 /m

Therefore,

1/λ = (1.097 x 10^7 /m)(1/1² - 1/∞²)

λ = 9.115 x 10^-8 m = 91.15 nm

Now, for energy (E) we know that:

E = hc/λ

where,

h = Plank's Constant = 6.625 x 10^-34 J.s

c = speed of light = 3 x 10^8 m/s

Therefore,

E = (6.625 x 10^-34 J.s)(3 x 10^8 m/s)/(9.115 x 10^-8 m)

E = (2.18 x 10^-18 J)(1 eV/1.6 x 10^-19 J)

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