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Anastaziya [24]
4 years ago
6

Two iron oxide samples are given to you where one is red and the other is black. You perform a chemical analysis and you find th

at the red sample has a Fe/O mass ratio of 2.327 and the black has a Fe/O mass ratio of 3.491. You suspect the red sample is simple rust or Fe2O3. What is the chemical formula for the black sample?
Chemistry
1 answer:
kupik [55]4 years ago
7 0

Answer:

Chemical formula for the black iron oxide sample is FeO

Explanation:

<u>Given:</u>

Mass ratio of black iron oxide sample, Fe/O = 3.491

<u>To determine:</u>

The chemical formula

<u>Calculation:</u>

The mass ratio is Fe:O = 3.491 : 1

Mass of Fe = 3.491 g

Mass of O = 1.000 g

Atomic mass of Fe =55.85 g/mol

Atomic mass of O = 16.00 g/mol

Moles\ Fe = \frac{3.491g}{55.85g/mol} =0.625\\\\Moles\ O = \frac{1.000g}{16.00g/mol} =0.0625\\

Therefore, the molar ratio of Fe:O = 1:1

Hence, chemical formula is FeO

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The synthesis of nitrogen trihydride from nitrogen gas and hydrogen gas is shown by which balanced chemical equation?
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 The correct answer is:  [B]:  
___________________________________________________________
                     "  N₂ (g) + 3 H₂ (g) → 2 NH₃ <span>(g)  "    .
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Note of interest:
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This particular reaction is known as the "</span>Haber process" .
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7 0
4 years ago
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A galvanic (voltaic) cell consists of an electrode composed of zinc in a 1.0 M zinc ion solution and another electrode composed
MariettaO [177]

Answer:

The E°cell for the galvanic cell is 1.56 V.

Explanation:

A galvanic cell is a device that uses redox reactions to convert chemical energy into electrical energy. The chemical reaction used is always spontaneous.

Oxide-reduction reactions, also called redox, involve the transfer or transfer of electrons between two or more chemical species. In these reactions two substances interact: the reducing agent and the oxidizing agent.

The gain of electrons is called reduction and the loss of electrons oxidation. That is to say, there is oxidation whenever an atom or group of atoms loses electrons (or increases its positive charges) and in the reduction an atom or group of atoms gains electrons, increasing its negative charges or decreasing the positive ones.

The species that supplies electrons is the reducing agent (that is, it is that species that oxidizes, yielding electrons and increasing its positive charge, or decreasing the negative one causing the reduction of the other species) and the one that gains them is the oxidizing agent ( that is, it is that species that is reduced, capturing electrons and increasing its negative charge, or decreasing its positive charge, causing oxidation of the other species).

The galvanic cell works as follows: In the anodic half-cell oxidations occur, while in the cathodic half-cell reductions occur. The anode electrode, conducts the electrons that are released in the oxidation reaction, to the metallic conductors. These electrical conductors conduct the electrons and carry them to the cathode electrode; the electrons thus enter the cathode half-cell and the reduction takes place in it.

To determine the oxidizing and reducing agent you must first know the reduction potentials. For this you consult the list of standard reduction potentials. In this list you can see that the semi-reactions that occur with their corresponding potentials are:

Ag⁺ + e⁻ ⇒ Ag E°= 0.80 V

Zn²⁺ + 2 e⁻ ⇒ Zn E° -0.76 V

The species that has the greatest potential for reduction will be the species that will be reduced, that is, it will be the oxidizing agent. In this case, it will be the experience corresponding to silver (Ag). Therefore, to obtain the redox reaction, the half-reaction corresponding to zinc (Zn) must be reversed to be an oxidation, keeping its E ° value constant. Then:

Reduction: Ag⁺ + e⁻ ⇒ Ag E°= 0.80 V

Oxidation: Zn ⇒ Zn²⁺ + 2 e⁻ E° -0.76 V

So: <em>E°cell=Ereduction - Eoxidation</em>

Or what is the same<em> E°cell=Ecathode - Eanode </em>because the reduction always occurs in the cathode and oxidation in the anode.

E°cell=0.80 V - (-0.76) V

<em>E°cell= 1.56 V</em>

Then <u><em>the E°cell for the galvanic cell is 1.56 V.</em></u>

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Apart from boiling point and flammability, what other quality does the size of the molecules in hydrocarbons affect?​
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Answer:

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Hope this helps!

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PLEASE HELP
antiseptic1488 [7]
I’m pretty sure it’s D. Hope that helps! Correct me if I am wrong.
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