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Elis [28]
3 years ago
13

Element x reacts with oxygen to produce x2o3 in an experiment it is found that 1.0000 g of x produces 1.1xxx g of x2o3 what is t

he molar mass of x?
Chemistry
1 answer:
8_murik_8 [283]3 years ago
4 0

Since X is 1 g, therefore O must be 0.1 g. Therefore:

moles O = 0.1 g / (16 g / mol) = 0.00625 mol

 

We can see that for every 3 moles of O, there are 2 moles of X, therefore:

moles X = 0.00625 mol O (3 moles X / 2 moles O) = 0.009375 mol

 

Molar mass X = 1 g / 0.009375 mol

<span>Molar mass X = 106.67 g/mol</span>

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If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?
Semenov [28]

Answer:

see explanation below

Explanation:

The question is incomplete. Here's the complete question:

<em>Consider the following reaction: </em>

<em>SO2Cl2 -----> SO2(g) + Cl2(g) </em>

<em>Kc= 2.99 x 10^-7 at 227 degrees celcius </em>

<em>If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?</em>

This is a problem of equilibrium, therefore, we need to solve this using the expression of equilibrium constant. To do that, we need to wirte an ICE chart and solve from there:

       SOCl2 ---------> SO2 + Cl2     Kc = 2.99x10⁻⁷

i)       0.168                  0        0

c)         -x                    +x       +x

e)     0.168-x                x         x

Writting the Kc expression:

Kc = [SO2] [Cl2] / [SOCl2]

Replacing the values from the chart:

2.99x10⁻⁷ = x² / 0.168 - x

However, Kc is a very very small value, therefore, we can assume that the value of "x" would be very small too, and we can neglect the 0.168-x and just round it to 0.168:

2.99x10⁻⁷ = x²/0.168

2.99x10⁻⁷ * 0.168 = x²

√5.02x10⁻⁸ = x

x = 2.24x10⁻⁴ M

This means then, that the concentration of Cl2 in equilibrium would be:

<em>[Cl₂] = 2.24x10⁻⁴ M</em>

5 0
3 years ago
This chart show the amount of radioactivity measured from three unknown isotopes.
Ad libitum [116K]

Answer:

We are asked to describe the half-life of the radioactive materials.

  • Hence the 3rd option is correct
5 0
3 years ago
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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A. conduction
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