a.
Acids react with bases and give salt and water and the products.
Hence, HCl reacts with NaOH and gives NaCl salt and H₂O as the products. The reaction is,
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
To balance the reaction equation, both sides hould have same number of elements.
Left hand side, Right hand side,
H atoms = 2 H atoms = 2
Cl atoms = 1 Cl atoms = 1
Na atoms = 1 Na atoms = 1
O atoms = 1 O atoms = 1
Hence, the reaction equation is already balanced.
b.
Molarity (M)= moles of solute (mol) / Volume of the solution (L)
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
Molarity of NaOH = <span>0.13 M
</span>Volume of NaOH added = <span>43.7 mL
Hence, moles of NaOH added = 0.13 M x 43.7 x 10</span>⁻³ L
= 5.681 x 10⁻³ mol
Stoichiometric ratio between NaOH and HCl is 1 : 1
Hence, moles of HCl = moles of NaOH
= 5.681 x 10⁻³ mol
5.681 x 10⁻³ mol of HCl was in <span>26.9 mL.
Hence, molarity of HCl = </span>5.681 x 10⁻³ mol / 26.9 x 10⁻³ L
= 0.21 M
We are given with
4.35 g Phosphoric acid
5.25 g KOH
3.15 g K3PO4 produced
The reaction is
H3PO4 + 3KOH => K3PO4 + 3H2O
First, convert masses into moles.
Then, determine the limiting reactant.
Next, determine the maximum amount of K3PO4 that can be produced from the limiting reactant.
Lastly, calculate the percent yield by dividing the actual amount produced by the theoretical amount produced.
Answer:
Increasing the volume of the vessel
Explanation:
By the Le Chatelier's principle, if a system in equilibrium suffer a variation that disturbs the equilibriu, the reaction shift in the way to minimize the pertubation and re-establish the equilibrium.
For a variation in pressure, when it increases, the reaction shift for the smallest of gas volume, and if decreases, the reaction will shift for the large gas volume. So, for the reaction given, the products have the large amount of gas, so by decreasing the pressure, more products will be formed, and the amount of NH₄HS will reduce. To decrease the pressure, we can increase the volume of the vessel: for the ideal gas equation (PV= nRT), pressure and volume are indirectly proportional.
Total mass of CaCO3 = 40 amu of Ca + 12amu of C + 16×3 amu of oxygen = 100amu of CaCO3
i.e 100 tonnes of CaCO3 .
mass of CO2 = 12amu of C + 2× 16amu of O = 44 amu of CO2
mass % of CO2 in CaCO3 = (44/100)×100 =44%
i.e
44% of 100 tonnes is CO2.
=44 tonnes of CO2.
therefore, 44% of CO2 is present in CaCO3.
Colligative properties are usually used in relation to solutions.
Colligative properties are those properties of solutions, which depend on the concentration of the solutes [molecules, ions, etc.] in the solutions and not on the chemical nature of those chemical species. Examples of colligative properties include: vapour pressure depression, boiling point elevation, osmotic pressure, freezing point depression, etc.
For the question given above, the correct option is D. This is because the statement is talking about freezing point elevation, which is not part of colligative properties.