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djyliett [7]
3 years ago
7

Consider the picture of a gas pump. Which type of gasoline has the highest percentage of octane (the main component of gasoline)

?
unleaded: 87 octane rating

mid-grade: 89 octane rating

premium: 91 octane rating
Chemistry
2 answers:
Readme [11.4K]3 years ago
5 0

Answer:

premium: 91 octane rating

Explanation:

Octane number refers to the percentage or volume fraction of isooctane in a fuel.

The octane number gives a picture of how safe a fuel is for an engine. The higher the octane rating the lesser the tendency of the fuel to cause knocking of the engine.

The type of gasoline with the highest percentage of octane among the options is premium.

riadik2000 [5.3K]3 years ago
4 0

Answer:

c. premium: 91 octane rating

Explanation:

i got it right on EDG2021

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A 100.0 mLflask is filled with 0.065 moles of A and allowed to react to form B according to the reaction below. The following ex
kodGreya [7K]

The question is incomplete. The complete question is :

A 100.0 mL flask is filled with 0.065 moles of A and allowed to react to form B according to the reaction below. The following experimental data are obtained for the amount of A as the reaction proceeds. What is the average rate of appearance of B in units of M/s between t = 10 min. and t = 30 min.? Assume that the volume of the flask is constant. A(g) → B(g)

Time 0.0 10.0 20.0 30.0 40.0

Moles of A 0.065 0.051 0.042 0.036 0.031

Solution :

Consider the following reaction as follows :

$A \rightarrow B$

The experiment data is given as follows :

Time (min) :   0.0        10.0       20.0        30.0       40.0

Moles of A :  0.065    0.051    0.042      0.036     0.031

According to the rate of reaction concept, the rate can be expressed as a consumption of the reactant and formation of the product as follows :

Average rate : $= -\frac{d[A]}{dt} =  \frac{d[B]}{dt} $

Now we have to calculate the average rate between 10.0 to 30.0 min w.r.t. A as follows :

Rate  $=-\frac{(0.051-0.036) mol \times \frac{1}{0.1 \ L}}{(30.0-10.0) mol \times \frac{60 \ s}{1 \ min}}$

         $=\frac{0.15 \ M}{20 \ min \times \frac{60 \ s}{1 \ min}}$

         $= 1.25 \times 10^{-4 }\ M/s$

Therefore, the rate = $= 1.3 \times 10^{-4 }\ M/s$

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