Answer:
Explanation:
The formula for the force exerted between two ions is
where the Coulomb constant
k = 8.988 × 10⁹ N·m·C⁻²
Data:
z₁ = +2; r₁ = 0.62 nm
z₂ = -1; r₂ = 0.121 nm
Calculations:
r = r₁ + r₂ = 0.62 + 0.121 = 0.741 nm
Answer:
Standard enthalpy of formation of Carbon disulfide CS2 = 87.3 KJ/mol
Explanation:
forming CS2 means that it should in the product side
C(graphite) + O2 → CO2 ΔH = -393.5
2S(rhombic) + 2O2 → 2SO2 ΔH = -296.4 x 2
CO2 + 2SO2 → CS2 + 3O2 ΔH = -1073.6 x -1
the second reaction is multiplied by 2 so that the SO2 and O2 can cancel out.
the third reaction is reversed (multiplied by -1) so that CS2 will be on the product side.
after adding the reaction and cancelling out similarities, the final reaction is: C(graphite) + 2S(rhombic) → CS2
Add ΔH to find the enthalpy of formation of CS2
ΔHf = (-393.5) + (-296.4 x 2) + (-1073.6 x -1) = 87.3 KJ/mol
be aware of signs
Answer:
63.02 g.
Explanation:
- Na reacts with Cl₂ according to the balanced equation:
<em>2Na + Cl₂ → 2NaCl,</em>
It is clear that 2 mole of Na react with 1 mole of Cl₂ to produce 2 moles of NaCl.
- Firstly, we need to calculate the no. of moles of Na and Cl₂:
no. of moles of Na = (mass/atomic mass) = (19.0 g/22.9897 g/mol) = 0.826 g.
no. of moles of Cl₂ = (mass/atomic mass) = (34.0 g/70.906 g/mol) = 0.48 g.
- From the stichiometry, Na reacts with Cl₂ with a molar ratio (2:1).
<em>So, 0.826 mol of Na "the limiting reactant" reacts completely with 0.413 mol of Cl₂ "left over reactant".</em>
The no. of moles of Cl₂ remained after the reaction = 0.48 mol - 0.413 mol = 0.067 mol.
∴ The mass of Cl₂ remained after the reaction = (no. of moles of Cl₂ remained after the reaction)(molar mass of Cl₂) = (0.067 mol)(70.906 g/mol) = 4.75 g.
- To get the no. of grams of produced NaCl:
<u><em>using cross multiplication:</em></u>
2 mol of Na produce → 2 mol of NaCl, from the stichiometry.
∴ 0.826 mol of Na produce → 0.826 mol of NaCl.
∴ The mass of NaCl produced after the reaction = (no. of moles of NaCl)(molar mass of NaCl) = (0.826 mol)(58.44 g/mol) = 48.27 g.
∴ The total weight of the glass vial containing the final product = the weight of the glass vial + the weight of the remaining Cl₂ + the weight of the produced NaCl = 10.0 g + 4.75 g + 48.27 g = 63.02 g.
