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2 years ago

Is iron likely to gain electrons, why?

Chemistry

1 answer:

Lana71 [14]2 years ago

7 0

Answer:

Explanation:

Fe (iron) loses three electrons.The ionic bond between ions results from the electrostatic attraction of opposite charges.

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Calculate no. Of moles in 329g of H2so4

Anestetic [448]

Mol= mass (grams) /Mr
Mr of Sulfuric Acid (H2SO4): 98
mol= 329/98
=3.36 moles

8 0

2 years ago

What is the partial pressure of argon, par, in the flask? express your answer to three significant figures and include the appro

monitta

Given which are missing in your question:
the flask is filled with 1.45 g of argon at 25 C°
So according to this formula (Partial pressure):
PV= nRT
first, we need n, and we can get by substitution by:
n = 1.45/mass weight of argon
= 1.45 / 39.948 = 0.0363 mol of Ar
we have R constant = 0.0821
and T in kelvin = 25 + 273 = 298
and V = 1 L
∴ P * 1 = 0.0363* 0.0821 * 298 = 0.888 atm

3 0

2 years ago

Standard reduction half-cell potentials at 25∘c half-reaction e∘ (v) half-reaction e∘ (v) au3 (aq) 3e−→au(s) 1.50 fe2 (aq) 2e−→f

Zanzabum

K = 5.0 × 10^25

<h2>Part (a). Calculate E° for the reaction </h2>

Step 1. Write the equations for the two half-reactions

2H^(+)(aq) + 2e^(-) → H2(g); _0.00 V

Zn^(2+)(aq) + 2e^(-) → Zn(s); -0.76 V

Step 2. Identify the cathode and the anode

The half-cell with the more negative E° (Zn) is the anode.

Step 3. Calculate E°

Zn(s) → Zn^(2+)(aq) + 2e^(-); _________+0.76 V

2H^(+)(aq) + 2e^(-) → H2(g); __________0.00 V

Zn(s) + 2H^(+)(aq) → Zn^(2+)(aq) + H2(g); +0.76 V

E° = +0.76 V

<h2>Part (b). Calculate K for the reaction </h2>

The relation between E° and K is

E° = (RT)/(nF)lnK

where

R = the universal gas constant: 8.314 J·K^(-1)mol^(-1)

T = the Kelvin temperature

n = the moles of electrons transferred

F = the Faraday constant: 96 485 J·V^(-1)mol^(-1)

Then

0.76 V = [8.314 J·K^(-1)mol^(-1) × 298.15 K]/[2 × 96 485 J·V^(-1)mol^(-1)]lnK

0.76 = 0.012 85 lnK

lnK = 0.76/0.012 85 = 59.16

K =e^59.16 = 5.0 × 10^25

4 0

3 years ago

Use the half-reaction method to balance the equation for the conversion of ethanol to acetic acid in acid solution:CH₃CH₂OH + Cr

garik1379 [7]

Balanced chemical equation is 3CH₃CH₂OH + 2Cr₂O₇²⁻ + 16H⁺ → 3CH₃COOH + 4Cr³⁺ + 11H₂O.

<h3>What is Balanced Chemical Equation ?</h3>

The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.

The chemical equation

CH₃CH₂OH + Cr₂O₇²⁻ → CH₃COOH + Cr³⁺

First assign the oxidation number for each atom in the equation.

$\overset{+3}{C}\overset{+1}{H_3} \overset{-1}{C} \overset{+1}{H_2} \overset{-2}{O} \overset{+1}{H} + \overset{+6}{Cr_2} \overset{-2}{O_7} + \overset{+1}{H} \rightarrow \overset{+3}{C}\overset{+1}{H_3} \overset{+3}{C}\overset{-2}{O} \overset{-2}{O} \overset{+1}{H} + \overset{+3}{Cr}$

Oxidation: C₂H₆O → C₂H₄O₂ + 4e⁻

Reduction: Cr₂O₇ + 6e⁻ → 2Cr⁺³

Now, balance the charge

Oxidation: C₂H₆O → C₂H₄O₂ + 4e⁻ + 4H⁺

Reduction: Cr₂O₇ + 6e⁻ + 14H⁺ → 2Cr⁺³

Now balance the oxygen atoms

Oxidation: C₂H₆O + H₂O → C₂H₄O₂ + 4e⁻ + 4H⁺

Reduction: Cr₂O₇ + 6e⁻ + 14H⁺ → 2Cr⁺³ + 7H₂O

Now, make electron gain equivalent to lost

Oxidation: C₂H₆O + H₂O → C₂H₄O₂ + 4e⁻ + 4H⁺ } × 3

Reduction: Cr₂O₇ + 6e⁻ + 14H⁺ → 2Cr⁺³ + 7H₂O } × 2

Now,

Oxidation: 3C₂H₆O + 3H₂O → 3C₂H₄O₂ + 12e⁻ + 12H⁺

Reduction: 2Cr₂O₇ + 12e⁻ + 28H⁺ → 4Cr⁺³ + 14H₂O

Now, add the both equations

3CH₃CH₂OH + 2Cr₂O₇²⁻ + 16H⁺ → 3CH₃COOH + 4Cr³⁺ + 11H₂O

Thus from the above conclusion we can say that the balanced chemical equation is 3CH₃CH₂OH + 2Cr₂O₇²⁻ + 16H⁺ → 3CH₃COOH + 4Cr³⁺ + 11H₂O.

Learn more about the Balanced chemical equation here: brainly.com/question/26694427

#SPJ4

7 0

1 year ago

All synthetic elements...

ki77a [65]

In chemistry, a synthetic element is a chemical element that does not occur naturally on Earth, and can only be created artificially. So far, 24 synthetic elements have been created (those with atomic numbers 95–118). All are unstable, decaying with half-lives ranging from 15.6 million years to a few hundred microseconds.

Here is a picture am if It doesn't show up, I am sorry

4 0

3 years ago