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Answer:
The muscular and nervous systems enable the involuntary breathing mechanism. The main muscles in inhalation and exhalation are the diaphragm and the intercostals (shown in blue), as well as other muscles. Exhalation is a passive action, as the lungs recoil and shrink when the muscles relax.
Explanation:
Answer:
The concentration of chloride ion is 
Explanation:
We know that 1 ppm is equal to 1 mg/L.
So, the 
content 100 ppm suggests the presence of 100 mg of in 1 L of solution.
The molar mass of is equal to the molar mass of Cl atom as the mass of the excess electron in is negligible as compared to the mass of Cl atom.
So, the molar mass of is 35.453 g/mol.
Number of moles = (Mass)/(Molar mass)
Hence, the number of moles (N) of present in 100 mg (0.100 g) of is calculated as shown below:
So, there is 
of present in 1 L of solution.
Answer:
the systematic name is called Lithium sulfide.
Answer:
a. Ksp = 4s³
b. 5.53 × 10⁴ mol³/dm⁹
Explanation:
a. Obtain an expression for the solubility product of AB2(S),in terms of s.
AB₂ dissociates to give
AB₂ ⇄ A²⁺ + 2B⁻
Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as
AB₂ ⇄ A²⁺ + 2B⁻
1 : 1 : 2
Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s
So, we have
AB₂ ⇄ A²⁺ + 2B⁻
[s] [s] [2s]
So, the solubility product Ksp = [A²⁺][B⁻]²
= (s)(2s)²
= s(4s²)
= 4s³
b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³
Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³
Substituting the value of s into the equation, we have
Ksp = 4s³
= 4(2.4 × 10³ mol/dm³)³
= 4(13.824 × 10³ mol³/dm⁹)
= 55.296 × 10³ mol³/dm⁹
= 5.5296 × 10⁴ mol³/dm⁹
≅ 5.53 × 10⁴ mol³/dm⁹
Ksp = 5.53 × 10⁴ mol³/dm⁹
