Answer: 
Explanation:
The balanced chemical equation :
To calculate the moles, we use the equation:

According to stoichiometry:
4 moles of
produces = 902.0 kJ of energy
415.1 moles of
produces =
of energy
Thus the change in enthalpy is 
Half-life<span> is the time required for a quantity to reduce to half its initial value. </span><span>If four half-lives have elapsed for calcium-45, then it would be 4x162.7 = 650.8 days have passed. Hope this answers the question. Have a nice day. Feel free to ask more questions.</span>
Answer:
What mass (g) of barium iodide is contained in 188 mL of a barium iodide solution that has an iodide ion concentration of 0.532 M?
A) 19.6
B) 39.1
C) 19,600
D) 39,100
E) 276
The correct answer to the question is
B) 39.1 grams
Explanation:
To solve the question
The molarity ratio is given by
188 ml of 0.532 M solution of iodide.
Therefore we have number of moles = 0.188 × 0.532 M = 0.100016 Moles
To find the mass, we note that the Number of moles =
from which we have
Mass = Number of moles × molar mass
Where the molar mass of Barium Iodide = 391.136 g/mol
= 0.100016 moles ×391.136 g/mol = 39.12 g
Flat as more oxygen and water can react over it think of it like this would a cube rust faster than a sheet
Answer:
Part A
K = (K₂)²
K = (K₃)⁻²
Part B
K = √(Ka/Kb)
Explanation:
Part A
The parent reaction is
2Al(s) + 3Br₂(l) ⇌ 2AlBr₃(s)
The equilibrium constant is given as
K = [AlBr₃]²/[Al]²[Br₂]³
2) Al(s) + (3/2) Br₂(l) ⇌ AlBr₃(s)
K₂ = [AlBr₃]/[Al][Br₂]¹•⁵
It is evident that
K = (K₂)²
3) AlBr₃(s) ⇌ Al(s) + 3/2 Br₂(l)
K₃ = [Al][Br₂]¹•⁵/[AlBr₃]
K = (K₃)⁻²
Part B
Parent reaction
S(s) + O₂(g) ⇌ SO₂(g)
K = [SO₂]/[S][O₂]
a) 2S(s) + 3O₂(g) ⇌ 2SO₃(g)
Ka = [SO₃]²/[S]²[O₂]³
[SO₃]² = Ka × [S]²[O₂]³
b) 2SO₂(g) + O₂(g) ⇌ 2 SO₃(g)
Kb = [SO₃]²/[SO₂]²[O₂]
[SO₃]² = Kb × [SO₂]²[O₂]
[SO₃]² = [SO₃]²
Hence,
Ka × [S]²[O₂]³ = Kb × [SO₂]²[O₂]
(Ka/Kb) = [SO₂]²[O₂]/[S]²[O₂]³
(Ka/Kb) = [SO₂]²/[S]²[O₂]²
(Ka/Kb) = {[SO₂]/[S][O₂]}²
Recall
K = [SO₂]/[S][O₂]
Hence,
(Ka/Kb) = K²
K = √(Ka/Kb)
Hope this Helps!!!