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Fiesta28 [93]
3 years ago
7

What is the answer to-6 + 1 = -5/

Chemistry
1 answer:
Jlenok [28]3 years ago
4 0
-6+1= -5
-5 is the true answer
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6. What are two uses for nonmetals?
lesantik [10]
Uses of nonmetals in our daily life: Oxygen which is 21% by volume helps in the respiration process.

Nonmetals used in fertilizers: Fertilizers contain nitrogen.

Nonmetals used in crackers: Sulphur and phosphorus are used in fireworks.
5 0
3 years ago
The coordination compound Co3[Cr(CN)6]2 contains Co2+ cations and a complex anion. What is the likely oxidation state for Cr in
otez555 [7]

Answer:

-3

Explanation:

The oxidation state or oxidation number of an atom is the total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom.

The complex anion here is [Cr(CN)6]3-.

Now, as the oxidation state of CN or cyanide ligand is -1, and if we suppose the oxidation state of Cr to be 'x', then; x - 6 = -3 (overall charge on the anion),

so x= +3. Hence the oxidation state of Chromium in this complex hexacyanochromium (III) anion comes out to be -3.

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7 0
3 years ago
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The initial concentrations of Pb2+ and Cu2+ are
VARVARA [1.3K]

Answer:

a) Ecell = 0.5123 V

b) Ecell =  0.4695 V

c) [Pb2 +] = 4.75 M

Explanation:

a)

The reaction at the cathode is represented as follows:

Cu2 + + 2e- -> Cu (s) Eocathode = 0.34 V

The reaction at the anode is equal to:

Pb (s) -> Pb2 + + 2e- Eoanode = -0.13 V

The number of moles of the electrons that are involved is equal to n = 2

Standard cell potential equals Eo = Eocathode - Eoanode = 0.34 V- (-0.13 V) = 0.47 V

 The initial cell potential can be calculated with the following formula:

Ecell = Eocell - - 0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V

b)

The reaction in the cell is equal to:

Cu2 + + Pb (s) -> Cu (s) + Pb2 +

The concentration of Cu2 that gives the exercise is equal 0.2 M

Therefore, the change in concentration for Cu2 + is equal to:

Cu2 + = 1.4 M - 0.2 M = 1.2 M

We use the formula from part a)

Ecell = Eocell - (0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V

c)

To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:

Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))

0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)

Clearing Pb2 +:

[Pb2 +] = 4.75 M

8 0
3 years ago
What is the most important role that water plays for living organisms?
Flauer [41]

Answer:

A. All organisms need water as energy source

8 0
3 years ago
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An element is?
Sidana [21]

i think it’s D , not sure though

4 0
3 years ago
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