Answer:
Recycling and reuse of materials
Explanation:
One of the greatest problems facing the human population is the problem of solid waste disposal. The menace of solid waste disposal has led to the idea of landfills. Land fills are depressions on the earth surface prepared for the purpose of solid waste disposal.
The most important approach towards solid waste disposal is the idea of recycling of materials. A material can be collected after use and processed into the same material or serve as a precursor in another manufacturing process. This means that no waste is generated as the materials which are supposed to be disposed of as solid waste are processed into other useful materials. This will reduce the volume of solid wastes generated that may need to be disposed in a landfill.
Answer : The final temperature of the copper is, 
Solution :
Formula used :
where,
Q = heat gained = 299 cal
m = mass of copper = 52 g
c = specific heat of copper = 
= final temperature = ?
= initial temperature = 
Now put all the given values in the above formula, we get the final temperature of copper.
Therefore, the final temperature of the copper is,
Mass of KNO₃ : = 40.643 g
<h3>Further explanation</h3>
Given
28.5 g of K₃PO₄
Required
Mass of KNO₃
Solution
Reaction(Balanced equation) :
2K₃PO₄ + 3 Ca(NO₃)₂ = Ca₃(PO₄)₂ + 6 KNO₃
mol K₃PO₄(MW=212,27 g/mol) :
= mass : MW
= 28.5 : 212,27 g/mol
= 0.134
Mol ratio of K₃PO₄ : KNO₃ = 2 : 6, so mol KNO₃ :
= 6/2 x mol K₃PO₄
= 6/2 x 0.134
= 0.402
Mass of KNO₃ :
= mol x MW KNO₃
= 0.402 x 101,1032 g/mol
= 40.643 g
Answer : The specific heat of tin is, 0.213 J/g.K
Explanation :
Formula used :
where,
q = amount of heat lost = -399.4 J
c = specific heat capacity of tin = ?
m = mass of tin = 25.0 g
= final temperature = 
= initial temperature = 
Now put all the given values in the above formula, we get:
Therefore, the specific heat of tin is, 0.213 J/g.K
Answer is: pH of barium hydroxide is 13.935.
Chemical dissociation of barium hydroxide in water:
Ba(OH)₂(aq) → Ba²⁺(aq) + 2OH⁻(aq).
c(Ba(OH)₂) = 0.43 M.
V(Ba(OH)₂) = 100 mL ÷ 1000 mL/L = 0.1 L.
n(Ba(OH)₂) = 0.43 mol/L · 0.1 L.
n(Ba(OH)₂) = 0.043 mol.
From chemical reaction: n(Ba(OH)₂) : n(OH⁻) = 1 : 2.
n(OH⁻) = 0.086 mol.
c(OH⁻) = 0.86 mol/L.
pOH = -logc(OH⁻).
pOH = 0.065.
pH = 14 - 0.065 = 13.935.
