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2 years ago

Is the electron configuration 1s22s22p5 for Oxygen correct?

Chemistry

1 answer:

joja [24]2 years ago

7 0

No that’s is the electronic configuration for fluorine.

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A certain half-reaction has a standard reduction potential +0.80 V . An engineer proposes using this half-reaction at the anode

Strike441 [17]

Answer:

a. Minimum 1.70 V

b. There is no maximum.

Explanation:

We can solve this question by remembering that the cell potential is given by the formula

ε⁰ cell = ε⁰ reduction - ε⁰ oxidation

Now the problem states the cell must provide at least 0.9 V and that the reduction potential of the oxidized species 0.80 V, thus

ε⁰ reduction - ε⁰ oxidation ≥ ε⁰ cell

Since ε⁰ oxidation is by definition the negative of ε⁰ reduction , we have

ε⁰ reduction - ( 0.80 V ) ≥ 0.90 V

⇒ ε⁰ reduction ≥ 1.70 V

Therefore,

(a) The minimum standard reduction potential is 1.70 V

(b) There is no maximum standard reduction potential since it is stated in the question that we want to have a cell that provides at leat 0.9 V

8 0

2 years ago

What is the charge of the nucleus in an Adam of oxygen -17

Ivenika [448]

Charge of nucleus is always positive whether it is element or isotope.

5 0

2 years ago

Let’s suppose I place a live mouse in closed chambers with a CO2 sensor. In Chamber A, the mouse was given caffeine; therefore t

adell [148]

Answer:

Explanation:

Chamber

6 0

3 years ago

Uranium-238 decays to lead-206 with a half-life of 4.5 x 109 yr. Determine how much uranium-238 decays in milligrams (to three s

Mamont248 [21]

Answer:

2.15 mg of uranium-238 decays

Explanation:

For decay of radioactive nuclide-

$N=N_{0}.(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}$

where N is amount of radioactive nuclide after t time, $N_{0}$ is initial amount of radioactive nuclide and $t_{\frac{1}{2}}$ is half life of radioactive nuclide

Here $N_{0}=4.60 mg$ , $t=4.1\times 10^{9}yr$ and $t_{\frac{1}{2}}=4.5\times 10^{9}yr$

So, $N=(4.60mg)\times (\frac{1}{2})^{\frac{4.1\times 10^{9}}{4.5\times 10^{9}}}$

so, N = 2.446 mg

mass of uranium-238 decays = (4.60-2.446) mg = 2.15 mg

3 0

3 years ago

Hydroxyl radicals react with and eliminate many atmospheric pollutants. However, the hydroxyl radical does not clean up everythi

Nesterboy [21]

Answer:

ΔHreaction = 263.15 kJ/mol

Explanation:

The reaction is as follow:

OH + CF₂Cl₂ → HOF + CFCl₂

You need to calculate the enthalpy of reaction and for this it is necessary to know the standard enthalpies for each of the compounds. These enthalpies are as follows and can be found in your textbook or on the Internet.

ΔHreaction = ∑ΔHproducts - ∑ΔHreactants

$delta(H)_{reaction} =((1*(-97.8)+(1*(-92))-((1*39)+(1*(-491.15))=263.15kJ/mol$

7 0

2 years ago