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kvv77 [185]
3 years ago
8

Calculate E o cell for the reaction: AgCl(s) + NO(g) → Ag(s) + Cl-(aq) + NO3-(aq) (in acidic solution) Use the fact that the red

uction potential for nitrate(aq) to give NO(g) is +0.96 V and for AgCl(s) it is +0.22 V. You should give your answer with 1 digit before the decimal point and to 2 decimal places. Do NOT include units..... Also - do not use scientific notation!
Chemistry
1 answer:
kaheart [24]3 years ago
7 0

Answer:

-0.74

Explanation:

Let's consider the following redox reaction.

AgCl(s) + NO(g) → Ag(s) + Cl⁻(aq) + NO₃⁻(aq)

We can identify the oxidation and reduction half-reactions.

Reduction (cathode): 1 e⁻ + AgCl(s) → Ag(s) + Cl⁻(aq)

Oxidation (anode): 2 H₂O(l) + NO(g) → NO₃⁻(aq) + 4H⁺(aq) + 3 e⁻

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

E° = E°red,cat - Ered,an = 0.22 V - 0.96 V = -0.74 V

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The piece of metal is magnetic
Alex_Xolod [135]

Answer:

Explanation:

Magnetic materials are always made of metal, but not all metals are magnetic. Iron is magnetic, so any metal with iron in it will be attracted to a magnet. Steel contains iron, so a steel paperclip will be attracted to a magnet too. Most other metals, for example aluminium, copper and gold, are NOT magnetic.

5 0
3 years ago
The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the r
rodikova [14]

Answer:

E₁ ≅ 28.96 kJ/mol

Explanation:

Given that:

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,

Let the activation energy for a catalyzed biochemical reaction = E₁

E₁ = ??? (unknown)

Let the activation energy for an uncatalyzed biochemical reaction = E₂

E₂ = 50.0 kJ/mol

    = 50,000 J/mol

Temperature (T) = 37°C

= (37+273.15)K

= 310.15K

Rate constant (R) = 8.314 J/mol/k

Also, let the constant rate for the catalyzed biochemical reaction = K₁

let the constant rate for the uncatalyzed biochemical reaction = K₂

If the  rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:

K₁ = 3.50 × 10³

K₂ = 1

Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;

we can use the formula for Arrhenius equation;

K=Ae^{\frac{-E}{RT}}

If K_1=Ae^{\frac{-E_1}{RT}} -------equation 1     &

K_2=Ae^{\frac{-E_2}{RT}} -------equation 2

\frac{K_1}{K_2} = e^{\frac{-E_1-E_2}{RT}

E_1= E_2-RT*In(\frac{K_1}{K_2})

E_1= 50,000-8.314*310.15*In(\frac{3.50*10^3}{1})

E_1 = 28957.39292  J/mol

E₁ ≅ 28.96 kJ/mol

∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol

8 0
3 years ago
14. What is the mass of 5.99 moles of C6H2C14.
Flauer [41]

Hey there!

Find molar mass of C₆H₂Cl₄.

C: 6 x 12.01

H: 2 x 1.008

Cl: 4 x 35.45

------------------

     215.876g

This is the mass of 1 mole of C₆H₂Cl₄.

Multiply 215.876 by 5.99.

215.876 x 5.99 = 1293

The mass of 5.99 moles of C₆H₂Cl₄ is 1293 grams.

Hope this helps!

6 0
4 years ago
What do are five differences for scientific method and engineering design process
seropon [69]
Science,math,art,technology,engineering
4 0
4 years ago
Cyanogen is a gas which contains 46.2% C and 53.8% N by mass. At a temperature of 25°C and a pressure of 750 mm Hg, 1.50 g of cy
sergiy2304 [10]

Answer:

Molecular formula of cyanogen is C₂N₂

Explanation:

We apply the ideal gases law to find out the mole of cyanogen

P . V =  n. R. T

Firstly let's convert the pressure in atm, for R

750 mmHg = 0.986 atm

25°C + 273 = 298K

0.986 atm . 0.714L = n . 0.082 L.atm/mol.K .298K

(0.986 atm . 0.714L) / (0.082 L.atm/mol.K .298K) = n

0.0288 mol = n

Molar mass of cyanogen = mass / mol

1.50 g /0.0288 mol = 52.02 g/m

Let's apply the percent, to know the quantity of atoms

100 g of compound contain 46.2 g of C and 53.8 g of N

52.02 g of compound contain:

(52.02 . 46.2) / 100 = 24 g  → 2 atoms of C

(52.02 . 53.8) / 100 = .28 g  →  2 atoms of N

3 0
3 years ago
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