Explanation:
Method of prepration of sodium thiosulphate - definition
In the laboratory, this salt can be prepared by heating an aqueous solution of sodium sulphite with sulphur or by boiling aqueous NaOH and sulfur according to this equation:
Opposite pairs form ionic bonds, due to this the answer is D - Li and Br; They have unlike charges.
Answer:
B. 0.5 molar
Explanation:
Given data:
Initial concentration = 0.40 M
Initial volume = 750 mL
Final volume =750 - 150 mL = 600 mL
Final concentration = ?
Solution:
Molarity is the number of moles of solutes in litter of solvent. In given problem it is stated that when the solution is uncovered solvent evaporate it means molarity is changed. we can calculate the new molarity with the following formula.
C₁V₁ = C₂V₂
C₁ = initial concentration
V₁ = initial volume
C₂ = final concentration
V₂ = final volume
Now we will put the values in formula.
C₁V₁ = C₂V₂
0.40 M × 750 mL = C₂ × 600 mL
300 M.mL / 600 mL = C₂
0.5 M = C₂
The concentration of AlCl3 solution if 150 ml of the solution contains 550 mg of cl- ion is 0.0344 M
calculation
concentration = moles /volume in liters
volume in liters = 150 /1000= 0.15 L
number of moles calculation
write the equation for dissociation of Al2Cl3
that is AlCl3 ⇔ Al^3+ + 3 Cl ^-
find the moles of Cl^- formed
moles =mass/molar mass
mass in grams= 550/ 1000 =0.55 grams
molar mass of Cl^- =35.5 g/mol
moles is therefore= 0.55/35.5 =0.0155 moles
by use of mole ration betweem AlCl3 to Cl^- which is 1:3 the moles of AlCl3 is =0.0155 x 1/3= 5.167 x10^-3 moles
concentration of AlCl3 is therefore= 5.167 x10^-3/ 0.15 =0.0344 M