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Annette [7]
2 years ago
10

On the basis of the information above, a buffer with a pH = 9 can best be made by using

Chemistry
1 answer:
telo118 [61]2 years ago
3 0

Answer:

D H2PO4– + HPO42–

Explanation:

The acid dissociation constant for \mathbf{H_3PO_4 , H_2PO^{-}_4 ,  HPO_4^{2-}} are \mathbf{7\times 10^{-3}, \ \ 8\times 10^{-8} ,\ \  5\times 10^{-13}} respectively.

\mathbf{pka (H_3PO_4) = -log (7\times 10^{-3} )=2.2}

\mathbf{pka (H_2PO_4^-) = -log (8\times 10^{-8} )=7.1}

\mathbf{pka (HPO_4^{2-}) = -log (5\times 10^{-13} )=12.3}

The reason while option D is the best answer is that, the value of pKa for both

\mathbf{H_2PO^{-}_4 ,\  \& \  HPO_4^{2-}} lies on either side of the desired pH of the buffer. This implies that one is slightly over and the other is slightly under.

Using Henderson-Hasselbach equation:

\mathbf{pH = pKa + log \Big( \dfrac{HPO_4^{2-}}{H_2PO_4^-} \Big)}

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5 0
3 years ago
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The Gateway Arch in St. Louis, MO is approximately 630 ft tall. How many U.S. dimes would be in a stack of the same
miskamm [114]

Answer:

142240

Explanation:

We are told in the question:

Height of Gateway Arch in St. Louis, MO = 630ft

We are asked, how many U.S. dimes would be in a stack of the same

height when 1 dime is 1.35 mm thick.

Step 1

Convert height in ft to mm

1 ft = 304.8 mm

630ft =

Cross Multiply

630ft × 304.8mm/1ft

= 192024 mm.

Step 2

To find how many US dimes would be in a stack of the same height

= Total thickness/ Thickness of 1 US dime

= 192024 mm/1.35mm

= 142240

Therefore, the number of dimes that would be in a stack of the same

height is 142240

6 0
3 years ago
Which of the following represent a mixture?
7nadin3 [17]
The answer is B NaCI solid
6 0
2 years ago
What are the two fundamental portions of an atom?​
Mice21 [21]
Number of Protons in an Uncharged Atom
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4 0
2 years ago
Combustion of hydrocarbons such as dodecane (C12H26) produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth'
miss Akunina [59]

Answer:

A. 2C12H26(l) + 37O2(g) —> 24CO2(g) + 26H2O(g)

B. 761.42 L

Explanation:

A. Step 1:

The equation for the reaction.

C12H26(l) + O2(g) —> CO2(g) + H2O(g)

A. Step 2:

Balancing the equation.

The equation can be balance as follow:

C12H26(l) + O2(g) —> CO2(g) + H2O(g)

There are 12 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 12 in front of CO2 as illustrated below:

C12H26(l) + O2(g) —> 12CO2(g) + H2O(g)

There are 26 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 13 in front of H2O as illustrated below:

C12H26(l) + O2(g) —> 12CO2(g) + 13H2O(g)

Now, there are a total of 37 atoms of O2 on the right side and 2 atoms on the left. It can be balance by putting 37/2 in front of O2 as illustrated below:

C12H26(l) + 37/2O2(g) —> 12CO2(g) + 13H2O(g)

Multiply through by 2 to clear the fraction from the equation.

2C12H26(l) + 37O2(g) —> 24CO2(g) + 26H2O(g)

Now the equation is balanced

B. Step 1:

We'll by obtaining the number of mole of C12H26 in 0.450 kg of C12H26. This is illustrated below:

Molar Mass of C12H26 = (12x12) + (26x1) = 144 + 26 = 170g/mol

Mass of C12H26 = 0.450 kg = 0.450x1000 = 450g

Number of mole of C12H26 =?

Number of mole = Mass/Molar Mass

Number of mole of C12H26 = 450/170

Number of mole of C12H26 = 2.65 moles

B. Step 2:

Determination of the number of mole of CO2 produced by the reaction. This is illustrated below:

2C12H26(l) + 37O2(g) —> 24CO2(g) + 26H2O(g)

From the balanced equation above,

2 moles of C12H26 produced 24 moles of CO2.

Therefore, 2.65 moles of C12H26 will produce = (2.65x24)/2 = 31.8 moles of CO2.

B. Step 3:

Determination of the volume of CO2 produced by the reaction.

Pressure (P) = 1 atm

Temperature (T) = 19°C = 19°C + 273 = 292K

Gas constant (R) = 0.082atm.L/Kmol

Number of mole (n) = 31.8 moles

Volume (V) =?

The volume of CO2 produced by the reaction can b obtained by applying the ideal gas equation as follow:

PV = nRT

1 x V = 31.8 x 0.082 x 292

V = 761.42 L

Therefore, the volume of CO2 produced is 761.42 L

5 0
3 years ago
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