Answer:
0.0344 M
Explanation:
- HC₂H₃O₂ (aq) + OH⁻ (aq) --> C₂H₃O₂⁻ (aq) + H₂O (l)
Because one mol of vinegar (acetic acid) reacts with one mol of NaOH, we can use the formula
Where C₁ and V₁ refer to the concentration and volume of vinegar, and C₂ and V₂ to those of NaOH. We're given V₁, C₂ and V₂; so we <em>solve for C₁</em>:
- C₁ = 0.0500 M * 16.7 mL / 25.0 mL
Answer : Both solutions contain
molecules.
Explanation : The number of molecules of 0.5 M of sucrose is equal to the number of molecules in 0.5 M of glucose. Both solutions contain
molecules.
Avogadro's Number is
=
which represents particles per mole and particles may be typically molecules, atoms, ions, electrons, etc.
Here, only molarity values are given; where molarity is a measurement of concentration in terms of moles of the solute per liter of solvent.
Since each substance has the same concentration, 0.5 M, each will have the same number of molecules present per liter of solution.
Addition of molar mass for individual substance is not needed. As if both are considered in 1 Liter they would have same moles which is 0.5.
We can calculate the number of molecules for each;
Number of molecules =
;
∴ Number of molecules =
which will be = 
Thus, these solutions compare to each other in that they have not only the same concentration, but they will have the same number of solvated sugar molecules. But the mass of glucose dissolved will be less than the mass of sucrose.
Answer:
[C] carbon solid
Explanation:
Pure solids and liquids are never included in the equilibrium constant expression because they do not affect the reactant amount at equilibrium in the reaction, thus since your equation has [C] as solid it will not be part of the equlibrium equation.
<u>Answer:</u> The chemical equations and equilibrium constant expression for each ionization steps is written below.
<u>Explanation:</u>
The chemical formula of carbonic acid is
. It is a diprotic weak acid which means that it will release two hydrogen ions when dissolved in water
The chemical equation for the first dissociation of carbonic acid follows:

The expression of first equilibrium constant equation follows:
![Ka_1=\frac{[H^+][HCO_3^{-}]}{[H_2CO_3]}](https://tex.z-dn.net/?f=Ka_1%3D%5Cfrac%7B%5BH%5E%2B%5D%5BHCO_3%5E%7B-%7D%5D%7D%7B%5BH_2CO_3%5D%7D)
The chemical equation for the second dissociation of carbonic acid follows:

The expression of second equilibrium constant equation follows:
![Ka_2=\frac{[H^+][CO_3^{2-}]}{[HCO_3^-]}](https://tex.z-dn.net/?f=Ka_2%3D%5Cfrac%7B%5BH%5E%2B%5D%5BCO_3%5E%7B2-%7D%5D%7D%7B%5BHCO_3%5E-%5D%7D)
Hence, the chemical equations and equilibrium constant expression for each ionization steps is written above.