Answer:
1) 4Fe + 3O2 → 2Fe2O3
2) H2 + Cl2 → 2HCl
3) 2Ag + H2S → Ag2S + H2
4) CH4 + 2O2 → CO2 + 2H2O
5) 2HgO → 2Hg + O2
6) 2Co + 3H2O → Co2O3 + 3H2
Answer:
pH after the addition of 10 ml NaOH = 4.81
pH after the addition of 20.1 ml NaOH = 8.76
pH after the addition of 25 ml NaOH = 8.78
Explanation:
(1)
Moles of butanoic acid initially present = 0.1 x 20 = 2 m moles = 2 x 10⁻³ moles,
Moles of NaOH added = 10 x 0.1 = 1 x 10⁻³ moles
CH₃CH₂CH₂COOH + NaOH ⇄ CH₃CH₂CH₂COONa + H₂O
Initial conc. 2 x 10⁻³ 1 x 10⁻³ 0
Equilibrium 1 x 10⁻³ 0 1 x 10⁻³
Final volume = 20 + 10 = 30 ml = 0.03 lit
So final concentration of Acid = 
Final concentration of conjugate base [CH₃CH₂CH₂COONa]
Since a buffer solution is formed which contains the weak butanoic acid and conjugate base of that acid .
Using Henderson Hasselbalch equation to find the pH
![pH=pK_{a}+log\frac{[conjugate base]}{[acid]} \\\\=-log(1.54X10^{-5} )+log\frac{0.03}{0.03} \\\\=4.81](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%2Blog%5Cfrac%7B%5Bconjugate%20base%5D%7D%7B%5Bacid%5D%7D%20%20%5C%5C%5C%5C%3D-log%281.54X10%5E%7B-5%7D%20%29%2Blog%5Cfrac%7B0.03%7D%7B0.03%7D%20%5C%5C%5C%5C%3D4.81)
Answer:
Oxygen is the limiting reactant.
Explanation:
Based on the reaction:
C₁₂H₂₂O₁₁ + 12O₂ → 12CO₂ + 11H₂O
<em>1 mole of sucrose reacts with 12 moles of oxygen to produce 12 moles of CO₂ and 11 moles of H₂O.</em>
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10.0g of sucrose (Molar mass: 342.3g /mol) are:
10.0g C₁₂H₂₂O₁₁ × (1mole / 342.3g) = 0.0292 moles of C₁₂H₂₂O₁₁
And moles of 10.0g of oxygen (Molar mass: 32g/mol) are:
10.0g O₂ × (1mole / 32g) = 0.3125 moles of O₂
For a complete reaction of 0.0292 moles of C₁₂H₂₂O₁₁ you need (knowing 12 moles of oxygen react per mole of sucrose):
0.0292 moles of C₁₂H₂₂O₁₁ × (12 moles O₂ / 1 mole C₁₂H₂₂O₁₁) = <em>0.3504 moles of O₂</em>
As you have just 0.3125 moles of O₂, <em>oxygen is the limiting reactant.</em>
Answer:
C. 1.17 grams
Explanation:
- The molarity is the no. of moles of solute in a 1.0 L of the solution.
<em>M = (mass/molar mass)solute x (1000/ V)</em>
M = 0.1 M, mass = ??? g, molar mass of NaCl = 58.44 g/mol, V = 200.0 mL.
∴ mass of NaCl = (M)(molar mass)(V)/1000 = (0.1 M)(58.44 g/mol)(200.0 mL)/1000 = 1.168 g ≅ 1.17 g.
