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Natalija [7]
3 years ago
12

A hydrogen fuel cell is more efficient than a combustion reaction because very little of the released energy is released as ___.

When used at home, a fuel cell eliminates the need for___
1) A. Heat
B. Electrical energy
C. Chemical energy

2) A. Oxygen
B. Hydrogen
C. An electrical grid
Chemistry
2 answers:
Studentka2010 [4]3 years ago
8 0
It would be A. heat and C. an electric grid.

Hope This Helps You!
Good Luck Studying :)
Klio2033 [76]3 years ago
7 0
A hydrogen fuel cell is more efficient than a combustion reaction because very little of the released energy is released as "Heat". <span>When used at home, a fuel cell eliminates the need for "An Electrical grid"

Hope this helps!</span>
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Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.45 M of reagent
sladkih [1.3K]

Answer:

\large\boxed{\large\boxed{0.0014M/s}}

Explanation:

From the table, first find the order of reaction, then find the rate constant, write the rate equation, and, finally, subsititue the data for the reaction that starts with 0.45M of reagent A and 0.90 M of reagents B and C.

<u>1. Table</u>

Trial  [A] (M)    [B] (M)   [C] (M)    Initial rate (M/s)

 1        0.20      0.20       0.20         6.0×10⁻⁵

2        0.20      0.20       0.60         1.8×10⁻⁴

3        0.40      0.20        0.20        2.4×10⁻⁴

4        0.40      0.40        0.20        2.4×10⁻⁴

<u>2. Orders</u>

a) From trials 3 and 4 you learn that the initial concentration of B does not change the change teh rate of the reaction. Hence the order with respect to [B] is 0.

b) From trials 1 and 2 you learn that when the concentration of C is tripled the rate of reaction is also tripled:

  • 0.60 / 0.2 = 3, and
  • 1.8×10⁻⁴ / 6.0×10⁻⁵ = 3

Hence, the order with respect to C is 1.

c) From trials 1 and 3 you get:

  • 0.40/0.2 = 2
  • 2.4×10⁻⁴ /  6.0×10⁻⁵ = 4

Which means that when the concentration of A is doubled, the rate of the reaction is quadruplicated. Hence, the order of the reaction with respect to A is 2.

<u>3. Rate equation</u>

Ther orders are:

              a=2\\\\b=0\\\\c=1

Hence the rate is:

            rate=k[A]^a{B}^b[C]^c\\ \\ rate=k[A]^2[B]^0[C]^1=k[A]^2C

<u>4. Rate constant, k</u>

<u />

You can use any trial to find the value of the constant, k

Using trial 1:

            6.0\times 10^{-5}M/s=k(0.20M)^2(0.20M)\\ \\ k=\frac{6.0\times 10^{-5}M/s}{(0.20M)^2(0.20M)}=0.0075M^{-2}s^{-1}

<u>5. Rate law:</u>

       rate=k[A]^2C=0.0075[A]^2[C]

<u>6. Substitute</u>

Subsititue the data for the reaction that starts with 0.45M of reagent A and 0.90 M of reagents B and C.

        rate=0.0075M^{-2}s^{-1}[A]^2[C]=0.0075M^{-2}s^{-1}[0.45M]^2[0.9M]

        r=0.00136688M/s\approx 0.0014M/s

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Number of electrons absorbed by the main energy level (n = 4):
VARVARA [1.3K]

Answer:

C

Explanation:

Because 2 x n square gives the total energy level been absorbed = 2 x16

=32

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