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stellarik [79]
2 years ago
9

Use enthalpies of formation given in appendix c to calculate δh for the reaction br2(g)→2br(g), and use this value to estimate t

he bond enthalpy d(br−br).
Chemistry
1 answer:
Contact [7]2 years ago
3 0

Given reaction represents dissociation of bromine gas to form bromine atoms

Br2(g) ↔ 2Br(g)

The enthalpy of the above reaction is given as:

ΔH = ∑n(products)ΔH^{0}f(products) - ∑n(reactants)ΔH^{0}f(reactants)

where n = number of moles

ΔH^{0}f= enthalpy of formation

ΔH = [2*ΔH(Br(g)) - ΔH(Br2(g))] = 2*111.9 - 30.9 = 192.9 kJ/mol

Thus, enthalpy of dissociation is the bond energy of Br-Br = 192.9 kJ/mol

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Which balanced redox reaction is occurring in the voltaic cell represented by the notation of A l ( s ) | A l 3 ( a q ) | | P b
frez [133]

The question is missing. Here is the complete question.

Which balanced redox reaction is ocurring in the voltaic cell represented by the notation of Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}?

(a) Al_{(s)}+Pb^{2+}_{(aq)} ->Al^{3+}_{(aq)}+Pb_{(s)}

(b) 2Al^{3+}_{(aq)}+3Pb_{(s)} -> 2Al_{(s)}+3Pb^{2+}_{(aq)}

(c)Al^{3+}_{(aq)}+Pb_{(s)} ->Al_{(s)}+Pb^{2+}_{(aq)}

(d) 2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

Answer: (d) 2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

Explanation: <u>Redox</u> <u>Reaction</u> is an oxidation-reduction reaction that happens in the reagents. In this type of reaction, reagent changes its oxidation state: when it loses an electron, oxidation state increases, so it is oxidized; when receives an electron, oxidation state decreases, then it is reduced.

Redox reactions can be represented in shorthand form called <u>cell</u> <u>notation,</u> formed by: <em><u>left side</u></em> of the salt bridge (||), which is always the <em><u>anode</u></em>, i.e., its half-equation is as an <em><u>oxidation</u></em> and <em><u>right side</u></em>, which is always <em><u>the cathode</u></em>, i.e., its half-equation is always a <em><u>reduction</u></em>.

For the cell notation: Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}

Aluminum's half-equation is oxidation:

Al_{(s)} -> Al^{3+}_{(aq)}+3e^{-}

For Lead, half-equation is reduction:

Pb^{2+}_{(aq)}+2e^{-} -> Pb_{(s)}

Multiply first half-equation for 2 and second half-equation by 3:

2Al_{(s)} -> 2Al^{3+}_{(aq)}+6e^{-}

3Pb^{2+}_{(aq)}+6e^{-} -> 3Pb_{(s)}

Adding them:

2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

The balanced redox reaction with cell notation Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)} is

2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

6 0
2 years ago
Sulfur is composed of three isotopes: 32S, 33S, and 34S. The atomic masses of these isotopes are given below. 32S: 31.97207 amu
elena-14-01-66 [18.8K]

Answer:

Abundance of 32S is 94.41%

Explanation:

The average atomic mass is defined as the sum of the atomic masses of each isotope times its abundance:

Average atomic mass = ∑ Atomic mass istope*Abundance

For the sulfur:

32.07amu = 31.97207X + 32.97146Y + 33.96786*0.0422 <em>(1)</em>

<em>Where X is abundance of 32S and Y abundance of 33S</em>

Also we can write:

1 = X + Y + 0.0422 <em>(2)</em>

0.9578 - X = Y

Because the sum of the abundances = 1

Replacing (2) in (1):

32.07amu = 31.97207X + 32.97146(0.9578 - X) + 33.96786*0.0422

32.07 = 31.97207X + 31.58006 - 32.97146X + 1.43344

-0.9435 = -0.99939X

0.9441  =X

In percentage, abundance of 32S is 94.41%

3 0
1 year ago
Identify the conjugate acid base pair <br> H3PO4(ag)+CO32=HCO3-(ag)+HPO42-(ag)
viktelen [127]

Answer:

H₃PO₄/H₂PO₄⁻ and HCO₃⁻/CO₃²⁻

Explanation:

An acid is a proton donor; a base is a proton acceptor.

Thus, H₃PO₄ is the acid, because it donates a proton to the carbonate ion.

CO₃²⁻ is the base, because it accepts a proton from the phosphoric acid.

The conjugate base is what's left after the acid has given up its proton.

The conjugate acid is what's formed when the base has accepted a proton.

H₃PO₄/H₂PO₄⁻ make one conjugate acid/base pair, and HCO₃⁻/CO₃²⁻ are the other conjugate acid/base pair.

H₃PO₄ + CO₃²⁻ ⇌ H₂PO₄⁻ + HCO₃⁻

acid       base         conj.       conj.

                               base       acid

3 0
2 years ago
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