To determine the number of moles of carbon dioxide that is produced, we need to know the reaction of the process. For the reaction of HCl and sodium carbonate, the balanced chemical equation would be expressed as:
2HCl + Na2CO3 = 2NaCl + H2O + CO2
From the initial amount given of sodium carbonate and the relation of the substances from the balanced reaction, we calculate the moles of carbon dioxide as follows:
0.2 moles Na2Co3 ( 1 mol CO2 / 1 mol Na2Co3 ) = 0.2 moles CO2
Therefore, the amount in moles of carbon dioxide that is produced from 0.2 moles sodium carbonate would be 0.2 moles as well.
Answer:
+3
Explanation:
The oxygen all have a -2 oxidation state. (peroxides are exceptions)
The chemical structure is symmetrical. Both carbon are equivalent.
2 (oxidation state of carbon) + 4 (oxidation state of oxygen) = charge of ion.
2 (oxidation state of carbon) + 4 (-2) = -2
oxidation state of carbon = +3
The reaction between NaOH and H₂SO₄ is as follows;
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of base to acid is 2:1
NaOH is a strong acid and H₂SO₄ is a strong acid, therefore complete ionization into their respective ions takes place.
number of acid moles reacted - 0.112 M / 1000 mL/L x 39.1 mL = 0.0044 mol
the number of base moles required for neutralisation = 0.0044 x 2 = 0.0088 mol
Number of NaOH moles in 25.0 mL - 0.0088 mol
Therefore in 1000 mL - 0.0088 mol/ 25.0 mL x 1000 mL/L = 0.352 mol/L
Therefore molarity of NaOH - 0.352 M
Answer:
See detailed mechanism in the image attached
Explanation:
The mechanism shown in detail below is the synthesis of serine in steps.
The first step is the attack of the ethoxide ion base on the diethyl acetamidomalonate substrate giving the enolate and formaldehyde.
The second step is the protonation of the oxyanion from (1) above to form an alcohol as shown.
Acid hydrolysis of the alcohol formed in (3) above yields a tetrahedral intermediate, a dicarboxyamino alcohol.
Decarboxylation of this dicarboxyamino alcohol yields serine, the final product as shown in the image attached.
