Answer:
1. Equivalence point
2. Direct titration
3. Primary standard
4. Titrand
5. Back titration
6. Standard solution
7. Titrant
8. Indirect titration
9. End point
10. Indicator
Explanation:
1. The equivalence point is the tiration point at which the quantity or moles of the added titrant is sufficient or equal to the quantity or moles of the analyte for the neutralization of the solution of the analyte.
2. Direct titration is a method of quantitatively determining the contents of a substance
3. A primary standard is an easily weigh-able representative of the mount of moles contained in a substance
4. A titrand is the substance of unknown concentration which is to be determined
5. The titration method that uses a given amount of an excess reagent to determine the concentration of an analyte is known as back titration
6. A standard solution is a solution of accurately known concentration
7. A titrant is a solution that has a known concentration and which is titrated unto another solution to determine the concentration of the second solution
8. Indirect titration is the process of performing a titration in athe reverse order
9. The end point is the point at which the indicator indicates that the equivalent quantities of the reagents required for a complete reaction has been added
10 An indicator is a compound used to visually determine the pH of a solution.
Answer:
The mass of Na₂O that can be produced by the chemical reaction of 4.0 grams of sodium with excess oxygen in the reaction is 5.39 grams.
Explanation:
You know the balanced reaction:
4 NA + O₂ ⟶ 2 Na₂O
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) react and are produced:
- Na: 4 moles
- O₂: 1 mole
- Na₂O: 2 moles
Being:
the molar mass of the compounds participating in the reaction is:
- Na: 23 g/mole
- O₂: 2*16 g/mole= 32 g/mole
- Na₂O: 2*23 g/mole +16 g/mole= 62 g/mole
Then by stoichiometry of the reaction they react and are produced:
- Na: 4 moles* 23 g/mole= 92 g
- O₂: 1 mole*32 g/mole= 32 g
- Na₂O: 2 moles* 62 g/mole= 124 g
Then you can apply the following rule of three: if 92 grams of Na produce 124 grams of Na₂O, 4 grams of Na, how much mass of Na₂O does it produce?
mass of Na₂O=5.39 g
<em><u>The mass of Na₂O that can be produced by the chemical reaction of 4.0 grams of sodium with excess oxygen in the reaction is 5.39 grams.</u></em>
Answer:
CO₃²⁻(aq) + 2H⁺(aq) → CO₂ (g) + H₂O (l)
Explanation:
The balanced reaction between Na2CO3 and HCl is given as;
Na₂CO₃ (aq) + 2 HCl (aq) → 2 NaCl (aq) + CO₂ (g) + H₂O (l)
The next step is o express the species as ions.
The complete ionic equation for the above reaction would be;
2Na⁺(aq) + CO₃²⁻(aq) + 2H⁺(aq) + 2Cl⁻(aq) → Na⁺(aq) + Cl⁻(aq) + CO₂ (g) + H₂O (l)
The next step is to cancel out the spectator ion ions; that is the ions that appear in both the reactant and product side unchanged.
The spectator ions are; Na⁺ and Cl⁻
The net ionic equation is given as;
CO₃²⁻(aq) + 2H⁺(aq) → CO₂ (g) + H₂O (l)
Answer:
Volume = 4.28L
Explanation:
Charles's gas law relates the volume and temperature of a gas at constant pressure. This law says that at constant pressures if we raise the temperature of a gas it will expand and if we reduce the temperature the volume will decrease. The formula is as follows:
So, the initial conditions are 2.22L and 23.9 °C ant the final conditions are 46.1 °C we replace them in the equation. And then we solve it.
