Answer:
The empirical formula is, C4H4S
Explanation:
Number of moles of carbon = 1.119 g/ 44g/mol = 0.025 moles
Mass of Carbon= 0.025 moles × 12 g/ mole = 0.3 g
Number of moles of hydrogen = 0.229/18g/mol × 2 = 0.025 moles
Mass of hydrogen = 0.025 moles × 1 = 0.025 g
Number of moles of sulphur = 0.407g/ 64 g/mol = 0.0064 moles
Mass of sulphur= 0.0064 moles ×32 = 0.2 g
Now we obtain the mole ratios by dividing through by the lowest ratio.
C- 0.025 moles/ 0.0064 moles, H- 0.025 moles/ 0.0064 moles, S- 0.0064 moles/0.0064 moles
C4H4S
At STP one mole of any gas occupies 22.4 L
moles NO2 = 99.0/22.4 = 4.42
mass NO2 = 4.42 mol x 46.0 g/mol=203 g
Answer:
Answer:
The mole ratio of C₄H₁₀ and CO₂ is 2 : 8, which simplifies to 1 : 4.
Explanation:
The mole ratio is the relative proportion of the moles of products or reactants that participate in the reaction according to the chemical equation.
The chemical equation given is:
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
Once you check that the equation is balanced, you can set the mole ratios for all the reactants and products. The coefficients used in front of each reactant and product, in the balanced chemical equation, tells the mole ratios.
In this case, they are: 2 mol C₄H₁₀ : 13 mol O₂ : 8 mol CO₂ : 10 mol H₂O
Since you are asked about the mole ratio of C₄H₁₀ and CO₂ it is:
2 mol C₄H₁₀ : 8 mol CO₂ , which dividing by 2, simplifies to
1 mol C₄H₁₀ : 4 mol CO₂, or
1 : 2.
Explanation:
Answer: -
Concentration of PbI₂ = 1.5 x 10⁻³ M
PbI₂ dissociates in water as
PbI₂ ⇄ Pb²⁺ + 2 I⁻
So PbI₂ releases two times the amount of I⁻ as it's own concentration when saturated.
Thus the molar concentration of iodide ion in a saturated PbI₂ solution = [ I⁻] =
= 1.5 x 10⁻³ x 2 M
= 3 x 10⁻³ M
PbI₂ releases the same amount of Pb²⁺ as it's own concentration when saturated.
[Pb²⁺] = 1.5 x 10⁻³ M
So solubility product for PbI₂
Ksp = [Pb²⁺] x [ I⁻]²
=1.5 x 10⁻³ x (3 x 10⁻³)²
= 4.5 x 10⁻⁹
