Metal ore has other elements in it as well. Also sediment and stone might cover the ore. We don't want to have a phone with sediment on it do we? thus these few reasons are why.
The rate equation is given as:
k = A e^(- Ea / RT)
Dividing state 1 and state 2:
k1/k2 = e^(- Ea / RT1) / e^(- Ea / RT2)
k1/k2 = e^[- Ea / RT1 - (- Ea / RT2)]
k1/k2 = e^[- Ea / RT1 + Ea / RT2)]
Taking the ln of both sides:
ln (k1/k2) = - Ea / RT1 + Ea / RT2
ln (k1/k2) = - Ea / R (1/T1 - 1/T2)
Since k2 = 4k1, therefore k1/k2 = ¼
ln (1/4) = [- (56,000 J/mol) / (8.314 J / mol K)] (1/273
K – 1/ T2)
2.058 x 10^-4 = 1/273 – 1/T2
T2 = 289.25 K
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Explanation:
The substance that is in excess that doesn't get used up as a reactant is called limiting reactant.
Answer:
The answer to your question is
4C₇H₁₇ + 45 O₂ ⇒ 28 CO₂ + 34H₂O
Explanation:
Write the equation
C₇H₁₇ + O₂ ⇒ CO₂ + H₂O
Process
1.- Check if the equation is balanced
Reactants Element Products
7 C 1
17 H 2
2 O 3
As the number of reactants and products is different, we conclude that the reaction is unbalanced.
2.- Write a coefficient "7" to CO₂ and a coefficient of 17/2 to H₂O
C₇H₁₇ + O₂ ⇒ 7CO₂ +
H₂O
Reactants Element Products
7 C 7
17 H 17
2 O 51/2
3.- Write a coefficient of 45/2 to the O₂, and multiply all the equation by 2.
4C₇H₁₇ + 45 O₂ ⇒ 28 CO₂ + 34H₂O
Reactants Element Products
28 C 28
68 H 68
90 O 90