The molecules are frozen in place but still vibrate
When sodium carbonate is dissolved in water, the equation is
.
When carbon dioxide is placed in water, aqueous carbon dioxide is formed: 
<h3>Dissolution of compounds in water</h3>
Some compounds are water-soluble, some are just partially soluble, while others are insoluble in water. Some soluble or partially soluble substances dissociate in water into their component ions. These substances are said to be ionic.
Sodium carbonate, like every other sodium salt, is soluble in water. It dissolves in water to form an aqueous solution of sodium carbonate.
While in solution, sodium carbonate dissociates into its component ions according to the following equation:

Carbon dioxide, on the other hand, does not dissociate in water. Instead, it dissolves in water where most of it remains as aqueous carbon dioxide in equilibrium with a small amount of hydronium ion and hydrogen carbonate ion.
Since the hydronium and hydrogen carbonate ions formed are so minute, the equation of the reaction can be written as: 
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Answer:
D. +5.7 kJ/mol
Explanation:
Molar free energy (ΔG) in the transportation of uncharged molecules as glucse through a cell membrane from the exterior to the interior of the cell is defined as:
ΔG = RT ln C in / C out
knowing R is 8,314472 kJ/molK; T is 298K Cin = 200mM and Cout = 20mM
ΔG = 5,7 kJ/mol
Right answer is:
D. +5.7 kJ/mol
I hope it helps!
The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH
<h3>What is Limiting reagent ?</h3>
The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.
Given chemical equation in balanced form ;
2NaOH(s) + CO₂(g) → Na₂CO₃(s) + H₂O(l).
According to the Chemical equation ;
- The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH
If 44 g CO₂ requires 80 g of NaOH, therefore, 4.4 g CO₂ will require atleast 8 g of NaOH.
But the available quantity is 5 g NaOH. thus, NaOH is the Limiting reagent.
- 6.625 g of Na₂CO₃ are expected to be produced 5.0 g of NaOH and 4.4 g of CO₂ are allowed to react
As 80 g NaOH produces 106 g of Na₂CO₃.
Therefore 5 g NaoH will produce ;
106 / 80 x 5 = 6.625 g
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