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Sergio039 [100]
3 years ago
12

Rhenium (Re) is made up of 37.4% 185Re and 62.6% 187Re. The atomic mass of 185Re is 184.953 amu. The atomic mass of 187Re is 186

.958 amu. What is the average atomic mass of rhenium?
Chemistry
2 answers:
Anna007 [38]3 years ago
8 0
To calculate the average mass of the element, we take the summation of the product of the isotope and the percent abundance. We calculate as follows:

Average atomic mass = .374 ( 184.953 amu ) + .626 ( <span>186.958 amu )  = 186.208 amu

Hope this answers the question. Have a nice day.</span>
Lelu [443]3 years ago
7 0

Answer:

186.207

Explanation:

done the topic

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A chemist designs a galvanic cell that uses these two half-reactions: half-reaction standard reduction potential (s)(aq)(aq)(l)
miv72 [106K]

Answer:

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)  

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻        

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

E°cell = 1.10 V

Explanation:

<em>The half-reactions are missing, but I will propose some to show you the general procedure and then you can apply it to your equations.</em>

<em>Suppose we have the following half-reactions.</em>

<em>Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)   E°red = 0.34 V</em>

<em>Zn²⁺(⁺aq) + 2 e⁻ → Zn(s)    E°red = -0.76 V</em>

<em />

To identify how to make a spontaneous cell, we need to consider the standard reduction potentials (E°red). The half-reaction with the higher E°red will occur as a reduction (in the cathode), whereas the one with the lower E°red will occur as an oxidation (in the anode).

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)   E°red = 0.34 V

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻        E°red = -0.76 V

To get the overall equation we add both half-reactions.

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E°cell = E°red, cat - E°red, an

E°cell = 0.34 V - (-0.76 V) = 1.10 V

Since E°cell > 0, the reaction is spontaneous.

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A solution is prepared by dissolving 20.1 g of MgCl2 with the addition of 157.0 mL of water (density of water is 1.00g/cm3). The
Nezavi [6.7K]

Answer:

1.3 M.

Explanation:

  • We need to calculate the mass of the solution:

mass of the solution = mass of MgCl₂ + mass of water

mass of MgCl₂ = 20.1 g.

mass of water = d.V = (157.0 mL)(1.0 g/cm³) = 157.0 g.

∴ mass of the solution = mass of MgCl₂ + mass of water = 20.1 g + 157.0 g = 177.1 g.

  • Now, we can get the volume of the solution:

V of the solution = (mass of the solution)/(density of the solution) = (177.1 g)/(1.089 g/cm³) = 162.62 mL = 0.163 L.

Molarity is the no. of moles of solute dissolved in a 1.0 L of the solution.

M = (no. of moles of MgCl₂) / (Volume of the solution (L)).

<em>∴ M = (mass/molar mass)of MgCl₂ / (Volume of the solution (L)) =</em> (20.1 g/95.211 g/mol) / (0.163 L) = <em>1.29 M ≅ 1.3 M.</em>

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Wofür steht die Abkürzung „PUR“?
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Explanation:

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