Almost all hydrocarbon 'burn' reactions involve oxygen; it's by far the most reactive substance in air.
<span>Hydrocarbon combustions always involve </span>
<span>[some hydrocarbon] + oxygen --> carbon dioxide + steam. </span>
C6H6(l) + O2 (g)--> CO2 (g)+ H2O (g)
<span>Balance carbon, six on each side: </span>
C6H6(l) + O2 (g)--> 6CO2 (g)+ H2O (g)
<span>Balance hydrogen, six on each side: </span>
C6H6(l) + O2 (g)--> 6CO2(g) + 3H2O (g)
<span>Now, we have fifteen oxygens on the right and O2 on the left. </span>
<span>Two ways to deal with that. We can use a fraction: </span>
C6H6 (l)+ (15/2)O2 (g)--> 6CO2 (g)+ 3H2O (g)
<span>Or, if you prefer to have whole number coefficients, double everything </span>
<span>to get rid of the fraction: </span>
2C6H6 (l)+ 15O2 (g)--> 12CO2 (g)+ 6H2O (g)
<span>With the SATP states thrown in... </span>
C6H6(l) + (15/2)O2(g) --> 6CO2(g) + 3H2O(g)
Answer:
The average rate is 2.84 X 10⁻³ Ms⁻¹
Explanation:
Average rate = -0.5*Δ[HBr]/Δt
given;
[HBr]₁ = 0.590 M
[HBr]₂ = 0.465 M
Δ[HBr] = [HBr]₂ - [HBr]₁ = 0.465 M - 0.590 M = -0.125 M
Δt Change in time = 22.0 s
Average rate = -0.5*Δ[HBr]/Δt
Average rate = - 0.5(-0.125)/22
Average rate = 0.00284 Ms⁻¹ = 2.84 X 10⁻³ Ms⁻¹
Therefore, the average rate is 2.84 X 10⁻³ Ms⁻¹
Answer:
sry i cant i to dubm need points tho
Explanation:
the answer is in the picture!
Answer: an amount of space between two things or people.
Explanation: hope this helps? :)
