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kramer
3 years ago
11

What is formed when water chemically combines with carbon dioxide

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

When water chemically combines with carbon dioxide, a Carbonic acid is formed.

<u>Explanation</u>:

  • Carbon dioxide responds with water in a solution to form a weak acid, carbonic acid. Carbonic acid disassociates into hydrogen particles and bicarbonate particles. The hydrogen particles and water respond with the most basic minerals modifying the minerals.  
  • Carbon dioxide and the other atmospheric gases disintegrate in surface waters. Dissolved gases are in equilibrium with the gas in the atmosphere. Carbon dioxide responds with water in a solution to form the weak acid, carbonic acid. Carbonic acid disassociates into hydrogen particles and bicarbonate particles.  
  • The hydrogen particles and water respond with the most basic minerals altering the minerals. The results of enduring are prevalently clays and soluble particles, for example, calcium, iron, sodium, and potassium. Bicarbonate particles additionally remain in solution; a remnant of the carbonic acid that was utilized to weather the rocks.

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A student wishes to calculate the experimental value of Ksp for AgI. S/he follows the procedure in Part 3 and finds Ecell to be
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Answer:

a)    [Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)    1.273 × 10⁻¹⁶

c)    2.629×10⁻¹⁹ M Thus; the value for  [Ag+ ]dilute will be too low

Explanation:

In an Ag | Ag+ concentration cell ,

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However, the Standard Reduction potential of cell = E°cell = 0

( since both cathode and anode have same Ag+║Ag )

Also , given that the theoretical slope is - 0.0591 V

Therefore; the reduction potential of cell ; i.e

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= 6.363 × 10⁻¹⁶ M × 0.20 M  

= 1.273 × 10⁻¹⁶

c) If s/he mistakenly uses 1.039 V as Ecell; then the value for [Ag+]dilute will be :

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

1.039 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 17.5804  

[Ag+]dilute = \mathbf{10^{-17.5804} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 2.629×10⁻¹⁹ M

Thus, the value for  [Ag+ ]dilute will be too low

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