There are 34 g of oxygen in the container.
We can use the<em> Ideal Gas Law</em> to solve this problem.
But 
, so
and
STP is 0 °C and 1 bar, so
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)
This is a double replacement because both Copper and Sodium reacted with each other compounds , hence "double"
Answer:
HI.
Explanation:
- Thomas Graham found that, at a constant temperature and pressure the rates of effusion of various gases are inversely proportional to the square root of their masses.
Rate of effusion ∝ 1/√molar mass.
- <em>(Rate of effusion of O₂) / (Rate of effusion of unknown gas) = (√molar mass of unknown gas) / (√molar mass of O₂).</em>
- An unknown gas effuses at one half the speed of that of oxygen.
∵ Rate of effusion of unknown gas = 1/2 (Rate of effusion of O₂)
∴ (Rate of effusion of O₂) / (Rate of effusion of unknown gas) = 2.
Molar mass of O₂ = 32.0 g/mol.
∵ (Rate of effusion of O₂) / (Rate of effusion of unknown gas) = (√molar mass of unknown gas) / (√molar mass of O₂).
∴ 2.0 = (√molar mass of unknown gas) / √32.0.
(
√molar mass of unknown gas) = 2.0 x √32.0
By squaring the both sides:
∴ molar mass of unknown gas = (2.0 x √32.0)² = 128 g/mol.
∴ The molar mass of sulfur dioxide = 80.91 g/mol and the molar mass of HI = 127.911 g/mol.
<em>So, the unknown gas is HI.</em>
<em></em>
Answer:
The valence electrons in the metal atom in the lattice structure can delocalize and move freely. The electrons are "loosely" connected with their parent atom in the metallic bond because most metals have excess electrons in their outer orbitals above the stable configuration.
Explanation:
Says google
