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jasenka [17]
3 years ago
6

Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced redox reaction at 25

°C.
2Al(s) + 3Mg2+ (aq) ® 2 Al3+(aq) + 3Mg(s)
Chemistry
1 answer:
o-na [289]3 years ago
6 0
The equation Eºcell = 0.0592/n logK must be used to find n and also Eºcell 
2 Al(s) + 3 Mg2+(aq) → 2 Al3+(aq) + 3 Mg(s) Al3+ +3e- --> Al Eº = -1.66 V Mg2+ +2e- -->Mg Eº = -2.37V 
To balance the equation, 6 moles of electrons must be transferred (2 Al and 3 Mg). This will be the value of n in the equation. 
To find Eºcell, you need the reduction potentials which should be given in a table, and given above. Eºcell = -1.66 - (-2.37) = 0.71 V log K = Eºcell x n/0.0592 = 0.71 x 6/0.0592 log K = 71.95 K = 10^71.95 K = 1.1x10^72
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6 0
3 years ago
A chemist must prepare of potassium hydroxide solution with a pH of at . He will do this in three steps: Fill a volumetric flask
maksim [4K]

The question is incomplete; the complete question is:

A chemist must prepare 800.0mL of potassium hydroxide solution with a pH of 13.00 at 25 degree C. He will do this in three steps: Fill a 800.0mL volumetric flask about halfway with distilled water. Weigh out a small amount of solid potassium hydroxide and add it to the flask. Fill the flask to the mark with distilled water. Calculate the mass of potassium hydroxide that the chemist must weigh out in the second step. Round your answer to 2 significant digits.

Answer:

4.5g (to 2 significant digits)

Explanation:

Now we must remember that KOH is a strong base, therefore it will practically dissociate completely.

To find the pH we can use the equation pH + pOH = 14.

Firstly to find the pOH:

13.00 + pOH = 14

pOH = 1.00

To find the [OH-]

Since

pOH= -log[OH^-]

[OH^-] = antilog (-pOH)

[OH^-]= antilog (-1)

[OH^-] = 0.1 molL-1

Since we've established that KOH is a strong base, we know that [OH-] = [KOH]

Also, we know that concentration = number of moles/volume

we have the concentration and the volume now so we can calculate the number of number of moles as follows:

The 800mL volume is the same as 0.8L

0.1 molL-1= number of moles/0.8L

0.08 moles = number of moles

now we can calculate the amount of solid KOH required

the molar mass of KOH = 39 + 16 +1 = 56 gmol-1

56 x 0.08 moles = 4.48g

So in 800mL of pH 13.00 KOH there is 4.5g of KOH dissolved.

3 0
3 years ago
What is the IUPAC name of 2NaOH(s)​
Feliz [49]

Answer:

NaoH= sodium hydroxide

6 0
3 years ago
The combustion of 3.795 mg of liquid B, which contains only C, H, and O, with excess oxygen gave 9.708 mg of CO2 and 3.969 mg of
attashe74 [19]

Answer:

Explanation:

9.708 mg of CO₂ will contain 12 x 9.708 / 44 = 2.64 g of C .

3.969 mg of H₂O will contain 2 x 3.969 / 18 = .441 g of H .

mg of O in given liquid B = 3.795 - ( 2.64 + .441 ) = .714 mg

ratio of mg of C , H , O in the compound = 2.64 : .441 : .714

ratio of no of atoms  of C , H , O in the compound

= 2.64 / 12 : .441 /1 : .714 / 16

= .22 : .44 : .0446

= .22 / .22 : .44 / .22 : .044 / .22

= 1 : 2 : .2

1 x 5 : 2 x 5 : .2 x 5

= 5 : 10 : 1

empirical formula of the compound = C₅H₁₀O

Volume of 89.8 mL at 1 .00 atm at 200⁰C

volume of gas at 1 atm and 0⁰C = 89.8 x 273 / 473 mL

= 51.83 mL

51.83 mL weighs .205 g

22400 mL will weigh .205 x 22400 / 51.83 g

= 88.6 g

So molecular weight = 88.6

Let molecular formula be (C₅H₁₀O)ₙ

molecular weight = n ( 5 x 12 + 10 + 16 )

= 86 n

86 n = 88.6

n = 1 approx

So molecular formula is same as empirical formula

C₅H₁₀O is molecular formula .

6 0
3 years ago
What is the hybridization of sb in the molecule sbcl52–? antimony (sb) can accept up to twelve electrons?
Setler79 [48]

The valence electron configuration for antimony (Sb) is:

Sb = 5s²5p³5d⁰

In SbCl₅²⁻, antimony has a -2 charge i.e. it has 2 additional electrons

Sb²⁻ = 5s²5p⁵5d⁰

Following a two electron transition from p→d orbital we have:

Sb²⁻ = 5s²5p³5d²

There is a total of 5 unpaired electrons (3 in the p and 2 in the d) which can form five bonds with the 5 Cl atoms.

Thus the hybridisation of Sb in SbCl₅²⁻ is sp³d²


5 0
3 years ago
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