Answer:
Zn°(s) + Fe⁺²(aq) => Zn⁺²(aq) + Fe°(s)
Explanation:
Molecular Equation:
Zn°(s) + Fe(NO₃)₂(aq) => Zn(NO₃)₂(aq) + Fe°(s)
Ionic Equation:
Zn°(s) + Fe⁺²(aq) + 2NO₃⁻(aq) => Zn⁺²(aq) + 2NO₃⁻(aq) + Fe°(s)
Net Ionic Equation: => Drop NO₃⁻ as spectator ion
Zn°(s) + Fe⁺²(aq) => Zn⁺²(aq) + Fe°(s)
Part A
75.0 mL of 0.10 M HF; 55.0 mL of 0.15 M NaF
This combination will form a buffer.
Explanation
Here, weak acid HF and its conjugate base F- is available in the solution
Part B
150.0 mL of 0.10 M HF; 135.0 mL of 0.175 M HCl
This combination cannot form a buffer.
Explanation
Here, moles of HF = 0.15 x 0.1 = 0.015 moles
Moles of HCl = 0.135 x 0.175 = 0.023
Since HCl is a strong acid and the number of HCl is higher than HF. This prevents the dissociation of HF and the conjugate base F- will not be available in the solution
Part C
165.0 mL of 0.10 M HF; 135.0 mL of 0.050 M KOH
This combination will form a buffer.
Explanation
Moles of HF = 0.165 x 0.1 = 0.0165 moles
Moles of KOH = 0.135 x 0.05 = 0.00675 moles
Moles of KOH is not sufficient for the complete neutralization of HF. Thus weak acid HF and its conjugate base F- is available in the solution and form a buffer
Part D
125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 M CH3NH3Cl
This combination will form a buffer
Explanation
Here, weak acid CH3NH3+ and its conjugate base CH3NH2 is available in the solution and form a buffer
Part E
105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl
This combination will form a buffer
Explanation
Moles of CH3NH2 = 0.105 x 0.15 = 0.01575 moles
Moles of HCl = 0.095 x 0.1 = 0.0095 moles
Thus the HCl completely reacts with CH3NH2 and converts a part of the CH3NH2 to CH3NH3+. This results weak acid CH3NH3+ and its conjugate base CH3NH2 is in the solution and form a buffer
Answer:
mass of U-235 = 15.9 g (3 sig. figures)
Explanation:
1 atom can produce -------------------------> 3.20 x 10^-11 J energy
x atoms can produce ----------------------> 1.30 x 10^12 J energy
x = 1.30 x 10^12 / 3.20 x 10^-11
x = 4.06 x 10^22 atoms
1 mol ----------------------> 6.023 x 10^23 atoms
y mol ----------------------> 4.06 x 10^22 atoms
y = 0.0675 moles
mass of U-235 = 0.0675 x 235 = 15.8625
mass of U-235 = 15.9 g (3 sig. figures)
It's a because if you add them together you till get 1.40
3 H2SO4 + 2 Al(OH)3 → Al2(SO4)3 + 6 H2O
(2.14 g Al(OH)3) / (78.0036 g Al(OH)3/mol) x (3 mol H2SO4 / 2 mol Al(OH)3) / (0.210 mol/L H2SO4) =
0.19596 L = 196 mL H2SO4
