Answer:
1.) 13 g C₄H₁₀
2.) 41 g CO₂
Explanation:
To find the mass of propane (C₄H₁₀) and carbon dioxide (CO₂), you need to (1) convert mass O₂ to moles O₂ (via molar mass), then (2) convert moles O₂ to moles C₄H₁₀/CO₂ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles C₄H₁₀/CO₂ to mass C₄H₁₀/CO₂ (via molar mass). It is important to arrange the ratios in a way that allows for the cancellation of units. The final answers should have 2 sig figs to match the sig figs of the given value.
Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)
Molar Mass (C₄H₁₀): 58.124 g/mol
Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)
Molar Mass (CO₂): 44.007 g/mol
Molar Mass (O₂): 2(15.998 g/mol)
Molar Mass (O₂): 31.996 g/mol
2 C₄H₁₀ + 13 O₂ ----> 8 CO₂ + 10 H₂O
48 g O₂ 1 mole 2 moles C₄H₁₀ 58.124 g
--------------- x ----------------- x -------------------------- x ------------------ =
31.996 g 13 moles O₂ 1 mole
= 13 g C₄H₁₀
48 g O₂ 1 mole 8 moles CO₂ 44.007 g
--------------- x ----------------- x -------------------------- x ------------------ =
31.996 g 13 moles O₂ 1 mole
= 41 g CO₂
Freezing point, boiling point, melting point, smell, attraction or repulsion to magnets, colour change, and many more examples.
Answer:
Kc = 4.774 * 10¹³
Explanation:
the desired reaction is
2 NO₂(g) ⇋ N₂(g) + 2 O₂(g)
Kc =[N₂]*[O₂]² /[NO₂]²
Since
1/2 N₂(g) + 1/2 O₂(g) ⇋ NO(g)
Kc₁= [NO]/(√[N₂]√[O₂]) → Kc₁²= [NO]²/([N₂][O₂])
and
2 NO₂(g) ⇋ 2 NO(g) + O₂(g)
Kc₂= [NO]²*[O₂]/[NO₂]² → 1/Kc₂= [NO₂]²/([NO]²[O₂])
then
Kc₁²* (1/Kc₂) = [NO]²/([N₂]*[O₂]) *[NO₂]²/([NO]²[O₂]) = [NO₂]²/([N₂]*[O₂]²) = 1/Kc
Kc₁² /Kc₂ = 1/Kc
Kc= Kc₂/Kc₁² =1.1*10⁻⁵/(4.8*10⁻¹⁰)² = 4.774 * 10¹³
Answer:
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