Answer:
1. Ba2+ 2. Sr2+
Explanation:
When a solution contains the Barium ,Ba²⁺ ion or Strontium, Sr²⁺ ion, they reacts with either H₂SO₄(aq) or Na₂SO₄(aq) to produce a white precipitate of BaSO₄(s) and SrSO₄(s) respectively
The chemical reactions are given below
Ba²⁺ + H₂SO₄(aq) ⇒ BaSO₄(s) + 2H⁺ (aq)
Ba²⁺ + Na₂SO₄(aq) ⇒ BaSO₄(s) + 2Na⁺ (aq)
Sr²⁺ + H₂SO₄(aq) ⇒ SrSO₄(s) + 2H⁺ (aq)
Sr²⁺ + Na₂SO₄(aq) ⇒ SrSO₄(s) + 2Na⁺ (aq)
Answer is: 1.29 grams <span>of solid formed.
</span>Chemical reaction: 2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq).
n(AgNO₃) = c(AgNO₃) · V(AgNO₃).
n(AgNO₃) = 0.220 M · 0.0351 L.
n(AgNO₃) = 0.0078 mol; limiting reactant.
n(K₂CrO₄) = 0.420 M · 0.052 L.
n(K₂CrO₄) = 0.022 mol.
From chemical reaction: n(AgNO₃) : n(Ag₂CrO₄) = 2 : 1.
n(Ag₂CrO₄) = 0.0078 mol ÷ 2.
n(Ag₂CrO₄) = 0.0039 mol.
m(Ag₂CrO₄) = 0.0039 mol · 331.73 g/mol.
m(Ag₂CrO₄) = 1.29 g.
Answer:
mass = 1.8x10⁻³ kg; number of moles = 4.1x10⁻⁵ kmol; specific volume = 0.55 m³/kg; molar specific volume = 24.4 m³/kmol
Explanation:
By the Avogadro's number, 1 mol of the matter has 6.02x10²³ molecules, thus, the number of moles (n) is the number of molecules presented divided by Avogadro's number:
n = 2.5x10²²/6.02x10²³
n = 0.041 mol
n = 4.1x10⁻⁵ kmol
The molar mass of CO₂ is 44 g/mol (12 g/mol of C + 2*16g/mol of O), and the mass is the number of moles multiplied by the molar mass:
m = 0.041 mol * 44 g/mol
m = 1.804 g
m = 1.8x10⁻³ kg
The specific volume (v) is the volume (1L = 0.001 m³) divided by the mass, and it represents how much volume is presented in each part of the mass:
v = 0.001/1.8x10⁻³
v = 0.55 m³/kg
The molar specific volume (nv) is the volume divided by the number of moles, and it represents how much volume is presented in each part of the mol:
nv = 0.001/4.1x10⁻⁵
nv = 24.4 m³/kmol
Nitrogen is the N and that is what is the element is the reactant in excess.
C. Homogeneous mixtures These alloys are homogeneous mixtures because they have a uniform composition throughout.
