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babymother [125]
2 years ago
11

Given that the rate constant is 4.0×10−4 m−1 s−1 at 25.0 ∘c and that the rate constant is 2.6×10−3 m−1 s−1 at 42.4 ∘c, what is t

he activation energy in kilojoules per mole? express your answer to three significant figures and include the appropriate units.
Chemistry
1 answer:
stepladder [879]2 years ago
4 0
So here's how you find the answer:

Given: (rate constants)

K₁ = 4.0 x 10⁻⁴ M⁻¹s⁻¹.
T₁ = 25.0 C = 293 K.
k₂ = 2.6 x10⁻³ M⁻¹s⁻¹.
T₂ = 42.4 C = 315.4 K.
R = 8,314 J/K·mol.

Use the equation:

ln(k₁/k₂) = Ea/R (1/T₂ - 1/T₁).

Transpose:

Ea = R·T₁·T₂ / (T₁ - T₂) · ln(k₁/k₂)

Substitute within the given transposed equation:

<span>Ea = 8,314 J/K·mol · 293 K · 315.4 K ÷ (293 K - 315.4 K) · ln (4.0 x 10⁻⁴ M⁻¹s⁻¹/ 2.6 x 10⁻³ M⁻¹s⁻¹).
</span>
Continuing the solution we get:

<span>Ea = 768315 J·K/mol ÷ (-22,4 K) * (-1.87)
</span>
The value of EA is:

<span>Ea = 64140.58 J/mol ÷ 1000 J/kJ = 64.140 kJ/mol.</span>


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Use the following half-reactions to design a voltaic cell: Sn4+(aq) + 2 e− → Sn2+(aq) Eo = 0.15 V Ag+(aq) + e−→ Ag(s) Eo = 0.80
AVprozaik [17]

Answer:

E° = 0.65 V

Explanation:

Let's consider the following reductions and their respective standard reduction potentials.

Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V

Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V

The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:

Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V

Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V

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

E° = E°red, cat - E°red, an = 0.80 V - 0.15 V = 0.65 V

4 0
2 years ago
Which of the following statements best describes the properties of aluminum-28?
Contact [7]

d. When aluminum-28 undergoes beta decay, silicon-28 is produced.

Explanation:

When the nuclei of aluminium-28 decays, it produces silicon- 28:

    Aluminium    ²⁸₁₃Al

    Silicon 28      ²⁸₁₄Si

    beta particle     ⁰₋₁\beta

       

           ²⁸₁₃Al        →          ²⁸₁₄Si      +        ⁰₋₁\beta

This way, the mass and atomic number are conserved.

Conservation of mass number:

                28 = 28 + 0,  28 = 28

                 13 = 14 -1 ,       13 = 13

Learn more:

Balancing nuclear equations brainly.com/question/10094982

#learnwithBrainly

       

5 0
2 years ago
7. You are about to perform some intricate electrical studies on single skeletal muscle fibers from a gastronemius muscle. But f
Vilka [71]

Answer:

58.61 grams

Explanation:

Taking The molecular weight of NaCl = 58.44 grams/mole

<u>Determine how many grams of NaCl to prepare the bath solution </u>

first we will calculate the moles of NaCl that is contained in 6L of 170 mM of NaCI solution

= ( 6 * 170 ) / 1000

= 1020 / 1000 = 1.020 moles

next

determine how many grams of NaCl

= moles of NaCl * molar mass of NaCl

= 1.020 * 58.44

= 58.61 grams

4 0
2 years ago
Water has unique properties which include its strength as a solvent; its three environmental stages of solid, liquid, and gas; a
pshichka [43]
Answer is: <span>unbalanced electronegativity of the hydrogens and oxygens as they share electrons.
Oxygen has greater electronegativity than hydrogen, because of that oxygen is partially negative and hydrogen is partially positive.
</span>Electronegativity<span> is a </span>chemical property<span> that describes the tendency of an </span>atom<span> to attract a shared pair of </span>electrons<span> towards itself.</span>
5 0
2 years ago
Determine the concentration of an aqueous solution that has an osmotic pressure of 3.9 atm at 37°C if the solute is sodium chlor
Snowcat [4.5K]

Answer:

Concentration for the solution is 0.153 mol/L

Explanation:

Formula for the osmotic pressure is  π = M . R . T . i

where M is molarity (concentration), R the universal constant for gases and T is Absolute T° (T°C + 273)

π = Osmotic pressure.

Let's replace the data given:

3.9 atm =  M . 0.082L.atm/mol.K . 310K

3.9 atm / 0.082 mol.K/L.atm . 310K = 0.153 mol/L (M)

i = Van't Hoff factor (ions from the solute dissolved in solution)

In this case, we assumed no ion pairing, so i = 1

8 0
2 years ago
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