Different amount of species in an area.
CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial Y - -
Change -X +X +X
Equilibrium Y-X X X
Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²
Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
X = 5.79 x 10⁻⁵ M
Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
= 5.79 x 10⁻⁵ mol/L
Molar mass of CaCO₃ = 100 g mol⁻¹
Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
= 5.79 x 10⁻³ g/L
Answer:
The answer to your question is letter B, 2-methylhexane.
Explanation:
Remember that for naming organic compounds first, we need to look for the largest chain of carbons.
In your example, the largest chain is horizontal and has 6 carbons.
Later, we need to circle all the branches, in your example there is only one branch located close to the left side
After that, we number the carbons of the main chain, starting in the corner with more branches, in your example we start from the first carbon on the left.
Finally, start naming the number of the carbon branch, later hte name of the branch and finally the name of the main chain.
Answer:
The four coefficients in order, separated by commas are 1, 8, 5, 6
Explanation:
We count the atoms in order to balance this combustion reaction. In combustion reactions, the products are always water and carbon dioxide.
C₅H₁₂ + ?O₂→ ?CO₂ + ?H₂O
We have 12 hydrogen in right side and we can balance with 6 in the left side. But the number of oxygen is odd. We add 2 in the right side, so we have 24 H, and in the product side we add a 12.
As we add 2 in the C₅H₁₂, we have 10 C, so we must add 10 to the CO₂ in the product side.
Let's count the oxygens: 20 from the CO₂ + 12 from the water = 32.
We add 16 in the reactant side. Balanced equation is:
2C₅H₁₂ + 16O₂→ 10CO₂ + 12H₂O
We also can divide by /2 in order to have the lowest stoichiometry
C₅H₁₂ + 8O₂→ 5CO₂ + 6H₂O
