Answer: 65.38g of Ca(OH)2 is needed
Explanation:
From The equation of reaction
2 HCl + Ca ( OH ) 2 ⟶ CaCl 2 + 2 H 2 O
NB: Molar mass of HCl= 1+35.5=36.5
Ca(OH)2= 74
From The stoichiometric equation
2mol of HCl(36.5×2=73) require 1mol of Ca(OH)2 (74g)
Hence 64.5g of HCl will require 64.5×74/73= 65.38g of Ca(OH)2
Answer:
Strontium
Explanation:
In the periodic table, an element with two (2) valence electrons is found on group 2. Group 2 is a group of the periodic table that harbors element called ALKALINE EARTH METALS. As the name implies, they are metals that possess shiny and solid characteristics at room temperature.
Group 2 elements include the following: Beryllium (Be), Magnesium (Mg), Calcium (Ca), Strontium (Sr), barium (Ba), and radium (Ra). Based on the descriptive information in this question, the element being described is a GROUP 2 element. Based on the elements in the option, only STRONTIUM (Sr) is a group 2 element.
Answer is: adding NaCl will lower the freezing point of a solution.
A solution (in this example solution of sodium chloride) freezes at a lower temperature than does the pure solvent (deionized water).
The higher the solute concentration (sodium chloride), freezing point depression of the solution will be greater.
Equation describing the change in freezing point:
ΔT = Kf · b · i.
ΔT - temperature change from pure solvent to solution.
Kf - the molal freezing point depression constant.
b - molality (moles of solute per kilogram of solvent).
i - Van’t Hoff Factor.
Dissociation of sodium chloride in water: NaCl(aq) → Na⁺(aq) + Cl⁻(aq).
Answer:
130ml of HCl(36%) in 4.90L solution => pH = 1.50
Explanation:
Need 4.90L of HCl(aq) solution with pH = 1.5.
Given pH = 1.5 => [H⁺] = 10⁻¹·⁵M = 0.032M in H⁺
[HCl(36%)] ≅ 12M in HCl
(M·V)concentrate = (M·V)diluted
12M·V(conc) = 0.032M·4.91L
=> V(conc) needed = [(0.032)(4.91)/12]Liters = 0.0130Liters or 130 ml.
Mixing Caution => Add 131 ml of HCl(36%) into a small quantity of water (~500ml) then dilute to the mark.
