1) Answer is: molar mas of ammonia is 17.031 g/mol.
M(NH₃) = Ar(N) + 3 · Ar(H) · g/mol.
M(NH₃) = 14.007 + 3 · 1.008 · g/mol.
M(NH₃) = 17.031 g/mol.
2) Answer is: molar mas of lead(II) chloride is 278.106 g/mol.
M(PbCl₂) = Ar(Pb) + 2 · Ar(Cl) · g/mol.
M(PbCl₂) = 207.2 + 2 · 35.453 · g/mol.
M(PbCl₂) = 278.106 g/mol.
3) Answer is: molar mas of acetic acid is 60.052 g/mol.
M(CH₃COOH) = 2 · Ar(C) + 2 · Ar(O) + 4 · Ar(H) · g/mol.
M(CH₃COOH) = 2 · 12.0107 + 2 · 15.9994 + 4 · 1.008 · g/mol.
M(CH₃COOH) = 60.052 g/mol.
Actually, no. While their mass may be the same (1kg), the volume of lead is a lot smaller than that of feathers. As there is the same mass stuffed in a smaller space, it must be denser. The density of water is 1 g/cm3, so if the density of the lead is more than 1g/cm3, it has to sink
Answer:
1.27 x 10⁻⁴M
Explanation:
For a 2 to 1 ionization ratio, solubility in pure water can be calculated using the formula S = ∛(Ksp/27) = ∛(5.61 x 10⁻¹¹/27 = 1.27 x 10⁻⁴M.
Mg(OH)₂ ⇄ Mg⁺² + 2OH⁻ => Ksp = [Mg⁺²][OH⁻]² = (x)(2x)² = 4x³
Solve for 'x' => x = Solubility = ∛Ksp/4
Explanation:
2CO (g) + O2 (g) → 2CO2 (g) ∆H = –566.0 kJ exothermic
2SO2(g) + O2(g) <=> 2SO3(g). Delta H = -198.2 kJ/mol. exothermic