Answer:
1.45 x 10⁻² g CO₂
Explanation:
To find the mass of carbon dioxide, you need to (1) convert grams CH₄ to moles CH₄ (via molar mass), then (2) convert moles CH₄ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams CO₂ (via molar mass). The final answer should have 3 sig figs to reflect the given value (5.30 x 10⁻³ g).
Molar Mass (CH₄): 12.011 g/mol + 4(1.008 g/mol)
Molar Mass (CH₄): 16.043 g/mol
Combustion of Methane:
1 CH₄ + 2 O₂ ---> 2 H₂O + 1 CO₂
Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)
Molar Mass (CO₂): 44.007 g/mol
5.30 x 10⁻³ g CH₄ 1 mole 1 mole CO₂ 44.007 g
--------------------------- x ---------------- x --------------------- x ----------------- =
16.043 g 1 mole CH₄ 1 mole
= 0.0145 g CO₂
= 1.45 x 10⁻² g CO₂
Adding the number of protons by the number of neutrons
The enthalpy change of the reaction below (ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
The bond energies data is given as follows:
BE for C≡O = 1072 kJ/mol
BE for Cl-Cl = 242 kJ/mol
BE for C-Cl = 328 kJ/mol
BE for C=O = 766 kJ/mol
The enthalpy change for the reaction is given as :
ΔHr×n = ∑H reactant bond - ∑H product bond
ΔHr×n = ( BE C≡O + BE Cl-Cl) - ( BE C=O + BE 2 × Cl-Cl )
ΔHr×n = ( 1072 + 242 ) - ( 766 + 656 )
ΔHr×n = 1314 - 1422
ΔHr×n = - 108 kJ
Thus, The enthalpy change of the reaction below ( ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
To learn more about enthalpy here
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Yes, depending on the thickness.
