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s2008m [1.1K]
2 years ago
15

The standard emf for the cell using the overall cell reaction below is +2.20 V:

Chemistry
1 answer:
ANEK [815]2 years ago
3 0

Answer:

emf generated by cell is 2.32 V

Explanation:

Oxidation: 2Al-6e^{-}\rightarrow 2Al^{3+}

Reduction: 3I_{2}+6e^{-}\rightarrow 6I^{-}

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

Overall: 2Al+3I_{2}\rightarrow 2Al^{3+}+6I^{-}

Nernst equation for this cell reaction at 25^{0}\textrm{C}-

E_{cell}=E_{cell}^{0}-\frac{0.059}{n}log{[Al^{3+}]^{2}[I^{-}]^{6}}

where n is number of electrons exchanged during cell reaction, E_{cell}^{0} is standard cell emf , E_{cell} is cell emf , [Al^{3+}] is concentration of Al^{3+} and [Cl^{-}] is concentration of Cl^{-}

Plug in all the given values in the above equation -

E_{cell}=2.20-\frac{0.059}{6}log[(4.5\times 10^{-3})^{2}\times (0.15)^{6}]V

So, E_{cell}=2.32V

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If 56 grams of notrogen are used up by the reaction, how many grams of amonia will be produced? 1N2+3H2--> 2NH3
djverab [1.8K]
1 mole of N2 produces 2 moles of NH3
OR...
14 x 2 grams of N2 produces 2(14 +3) grams of NH3
1 gram of N2 produces 34/28 grams of NH3
therefore, 56 grams produce (34/28 )x 56 =68 grams of NH3 

the answer thus would be 68 grams of NH3
5 0
3 years ago
A phosphate buffer is involved in the formation of urine. The developing urine contains H2PO4 and HPO42- in the same concentrati
Katen [24]

Answer:

The ionization equation is

H_{2}PO_{4}^{-}  +H_{2}O ⇄HPO_{4}^{-2}  +H_{3}O^{+} (1)

Explanation:

The ionization equation is

H_{2}PO_{4}^{-}  +H_{2}O ⇄HPO_{4}^{-2}  +H_{3}O^{+} (1)

As the Bronsted definition sais, an acid is a substance with the ability to give protons thus, H2PO4 is the acid and HPO42- is the conjugate base.

The Ka expression is the ratio between the concentration of products and reactants of the equilibrium reaction so,

Ka = \frac{[HPO_{4}^{-2}] [H_{3}O^{+}]}{[H_{2}PO_{4}^{-}] [H_{2}O]} = 6.2x10^{-8}

The pKa is

-Log (Ka) = -Log (6.2x10^{-8}) = 7.2

The pKa of H2CO3 is 6,35, thus this a stronger acid than H2PO4. The higher the pKa of an acid greater the capacity to donate protons.

In the body H2CO3 is a more optimal buffer for regulating pH due to the combination of the two acid-base equilibriums and the two pKa.

If the urine is acidified, according to Le Chatlier's Principle the equilibrium (1)  moves to the left neutralizing the excess proton concentration.

3 0
3 years ago
Assuming both graduated cylinders are holding water at room temperature, which cylinder has more thermal energy?
Serjik [45]

Answer:

D

Explanation:

bothe cylinders are at room temperature so no cylinder has more themal energy. That crosses off the answers A B and C.

5 0
3 years ago
A 17.4 L sample of oxygen gas (O2) was collected at a temperature of 23.0°C and a pressure of 2.18 atmospheres. What volume woul
timofeeve [1]

The volume of the gas at STP = 35.01 L

<h3>Further explanation</h3>

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure).

In general, the gas equation can be written  

\large {\boxed {\bold {PV = nRT}}}

where  

P = pressure, atm  

V = volume, liter  

n = number of moles  

R = gas constant = 0.08206 L.atm / mol K  

T = temperature, Kelvin  

V=17.4 L

T = 23 + 273 = 296 K

P = 2.18 atm

\tt mol=n=\dfrac{PV}{RT}\\\\n=\dfrac{2.18\times 17.4}{0.082\times 296}\\\\n=1.563

The volume of the gas occupy at STP :

\tt 1.563\times 22.4=35.01~L

6 0
2 years ago
if 980 KJ of energy are added to 6.2 L of water at 291 K, what will the final temperature of the water be
uranmaximum [27]

Answer:

The final temperature of the water will  be 328.81 K .

Explanation:

Using the equation, <em>q = mcΔT</em>

here, <em>q = energy</em>

<em>m= mass</em>

<em>c= specific heat capacity </em>

<em>ΔT= change in temperature</em>

<em>Mass of water = 1kg (1000 g ) per liter</em>

<em>∴ 6.2 Liter of water = 6200 g</em>

<em>c of water ≈ 4.18 J /g/K</em>

<em>Now, </em>

<em>980000 = 6200*4.18*ΔT</em>

<em>ΔT = 37.81 K</em>

<em>∴ final temperature of the water = 291 + 37.81 = 328.81 K</em>

6 0
3 years ago
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