Answer:
1.76 g/mL
Explanation:
You need to find the volume. You can do this by subtracting the volume of the water and the rock by the volume of the water.
72.7 mL - 50 mL = 22.7 mL
Now that you have volume, divide the mass by the volume to find the density.
39.943 g/22.7 mL = 1.76 g/mL
Answer is: pH of solution is 5,17.
Kb(NH₃) = 1,8·10⁻⁵.
c(NH₄Cl) = 0,084 M = 0,084 mol/L.
Chemical reaction: NH₄⁺ + H₂O → NH₃ + H₃O⁺.
Ka · Kb = 10⁻¹⁴.
Ka(NH₄⁺) = 10⁻¹⁴ ÷ 1,8·10⁻⁵.
Ka(NH₄⁺) = 5,55·10⁻¹⁰.
[H₃O⁺] = [NH₃] = x.
Ka(NH₄⁺) = [H₃O⁺] · [NH₃] ÷ [NH₄⁺].
5,55·10⁻¹⁰ = x² ÷ (0,084 M - x).
Solve quadratic equation: x = [H₃O⁺] = 6,8·10⁻⁶ M.
pH = -log[H₃O⁺].
pH = -log(6,8·10⁻⁶ M) = 5,17.
probably none becuase it going staright unless its going down a hill
Answer:
1 = oxidation
2 = reduction
Explanation:
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
2I- ----> I₂+ 2e⁻
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
F + e⁻ ----> F⁻
Consider the following reactions.
4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl
the oxidation state of copper is changed from +2 to +1 so copper get reduced.
CO + H₂O → CO₂ + H₂
the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.
Na₂CO₃ + H₃PO₄ → Na₂HPO₄ + CO₂ + H₂O
The oxidation state of carbon on reactant side is +4. while on product side is also +4 so it neither oxidized nor reduced.
H₂S + 2NaOH → Na₂S + 2H₂O
The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.
Answer:
0.928 M
Explanation:
The concentration of acid can be determined by using the volume used and the concentration and volume used of base.
We will use the law of equivalence of moles.
M₁V₁=M₂V₂
M₁ = concentration of base used
V₁ = volume of base used
M₂ = concentration of acid used =? (to be determined)
V₂ = volume of acid used
The initial concentration of KOH used is diluted so let us find the final concentration of KOH after dilution
initial moles = final moles
initial concentration X initial volume = final concentration X final volume
6.2 X 2.1 = 250 X final concentration
final concentration = 0.052 M = M₁
V₁ = 36.9 mL
V₂ = 6.2 mL
Here with each mole of phosphoric acid three moles of KOH are used.
Therefore
3 M₁V₁ = M₂V₂
M₂ = 