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irga5000 [103]
3 years ago
12

Calculate the standard cell potential (E∘) for the reaction X(s)+Y+(aq)→X+(aq)+Y(s) if K = 5.59×10−3.

Chemistry
1 answer:
Blizzard [7]3 years ago
5 0

Answer:

E° = -0.133 V

Explanation:

In the reaction:

X(s) + Y⁺(aq) → X⁺(aq) + Y(s)

<em>1 electron is transferred from X to Y</em>

Now, using Nernst equation:

E° = RT / nF ln K

<em>Where R is gas constant (8.314 J/molK), T is absolute temperature (Usually 298.15K), n are transferred electrons (1, for the reaction), F is faraday constant (96485C/mol) and K is equilibrium constant (5.59x10⁻³)</em>

Replacing:

E° = 8.314 J/molK*298.15K / 96485C/mol*1 ln 5.59x10⁻³

<em>E° = -0.133 V</em>

You might be interested in
determine the ph of a buffer that is 0.55 M HNO2 and 0.75 M KNO2. tha value of Ka for HNO2 is 6.8*10^-4
Mariana [72]

Answer:

pH = 3.3

Explanation:

Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.

In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].

Solution using the I.C.E. table:

              HNO₂ ⇄    H⁺   +   KNO₂⁻

C(i)        0.55M       0M      0.75M

ΔC            -x            +x          +x

C(eq)  0.55M - x       x     0.75M + x    b/c [HNO₂] / Ka > 100, the x can be                                    

                                                             dropped giving ...

           ≅0.55M        x       ≅0.75M        

Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]

=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M

pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3

Solution using the Henderson-Hasselbalch Equation:

pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]

= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]

= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3

3 0
3 years ago
Please answer fast, I will give brainliest if correct!
gizmo_the_mogwai [7]

Answer:

Explanation: 30 min

5 0
2 years ago
Under the right conditions aluminum will react with chlorine to produce aluminum chloride.
salantis [7]

Answer:

m_{Al}=9.42gAl

Explanation:

Hello there!

In this case, according to the given chemical reaction:

2 Al + 3 Cl2 --> 2 AlCl3

Whereas there is a 2:3 mole ratio of aluminum to chlorine; it will be possible for us to calculate the required grams of aluminum by using the equality 22.4 L = 1 mol, the aforementioned mole ratio and the atomic mass of aluminum (27.0 g/mol) to obtain:

m_{Al}=11.727LCl_2*\frac{1molCl_2}{22.4LCl_2}*\frac{2molAl}{3molCl_2}  *\frac{27.0gAl}{1molAl} \\\\m_{Al}=9.42gAl

Regards!

8 0
2 years ago
Calculate the frequency
Ber [7]
Given: wavelength of Nitrogen laser (∧) = 337.1 nm = 337.1 X 10^-9 m

We know that, Energy of photon (E) = hc/∧ = hv
where, v = frequency of photon and c = speed of light = 3 X 10^8 m/s

Thus, v = c/∧ = (3 X 10^8)/ (337.1 X 10^-9) = 8.899 X 10^14 s-1.

Answer: F<span>requency of nitrogen laser = </span>8.899 X 10^14 s-1.
3 0
3 years ago
Read 2 more answers
Refer to the following compounds.
goblinko [34]

Answer: The answer is D. This has a Carboxylic Acid group, and is acetic acid, or Ethanoic Acid.

ALWAYS LOOK for the Functional Group in question.

A. Would likely not stay in water, or at least not be acidic, for it is butane gas.

B. Is 1-propanol, and alcohols are not acidic as a rule. Certainly not in water.

C. This is an Ether. It will not give up an H+, it it not an acid.

E. This functional group is an amine, which is more “base” like, since the lone pairs of the Nitrogen atom would tend to attract a H+.

5 0
3 years ago
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