Answer:
The formula of the original halide is SrCl₂.
Explanation:
- The balanced equation of this reaction is:
SrX₂ + H₂SO₄ → SrSO₄ + 2 HX, where X is the halide.
- From the equation stichiometry, 1.0 mole of strontium halide will result in 1.0 mole of SrSO₄.
- The number of moles of SrSO₄ <em>(n = mass/molar mass) </em>= (0.755 g) / (183.68 g/mole) = 4.11 x 10⁻³ mole.
- The number of moles of SrX are 4.11 x 10⁻³ moles from the stichiometry of the balanced equation.
- n = mass / molar mass, n = 4.11 x 10⁻³ moles and mass = 0.652 g.
- The molar mass of SrX₂ = mass / n = (0.652) / (4.11 x 10⁻³ moles) = 158.62 g/mole.
- The molar mass of SrX₂ (158.62 g/mole) = Atomic mass of Sr (87.62 g/mole) + (2 x Atomic mass of halide X).
- The atomic mass of halide X = (158.62 g/mole) - (87.62 g/mole) / 2 = 71 / 2 g/mole = 35.5 g/mole.
- This is the atomic mass of Cl.
- <em>So, the formula of the original halide is SrCl₂</em>.
Answer: B
Explanation: molarity = concentration c= n/V = 0.5 mol/ 0.05 l = 10 mol/l
Answer:
Four electrons are present in the valence shell of Silicon.
Explanation:
Valence shell electrons are those electrons which are present in the outermost shell of an atom. These valence shell electrons are responsible for in the formation of bonds with other atoms.
Silicon having atomic number 14 has fourteen electrons in its neutral state and has the electronic configuration as follow;
1s², 2s², 2p⁶, 3s², 3p²
In given configuration the valence shell (outermost shell) is 3 and the number of electrons present in it are four i.e. 3s² and 3p² (2 + 2 = 4) respectively.
Answer:
S = 7.9 × 10⁻⁵ M
S' = 2.6 × 10⁻⁷ M
Explanation:
To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0
C +S +S
E S S
The solubility product (Ksp) is:
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²
S = 7.9 × 10⁻⁵ M
<u>Solubility in 0.0120 M CoBr₂ (S')</u>
First, we will consider the ionization of CoBr₂, a strong electrolyte.
CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)
1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.
Then,
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0.0240
C +S' +S'
E S' 0.0240 + S'
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')
In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.
S' = 2.6 × 10⁻⁷ M
When an atom combines chemically with another atom, it either gains, loses, or shares ELECTRONS.
