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Lapatulllka [165]
3 years ago
15

During a running (dynamic) compression test at idle, a good engine should produce PSI. A. 150 to 200 B. 100 to 150 C. 30 to 60 D

. 60 to 90​
Chemistry
1 answer:
Mazyrski [523]3 years ago
7 0

Answer:

b

Explanation:

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the anserw should be 665KJ

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kodGreya [7K]

Answer:It is C i got it correct

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In two or more complete sentences explain how to balance the chemical equation and classify its reaction type
Dafna11 [192]

Answer:

Hi! In this case, the reactioncan be correctly balance according to this: 2Al(s) + 3CuSO4(aq) –> Al2(SO4)3(aq) + 3Cu(s).

Explanation:

In this particulary reaction,two semi-reactions happens.

One involving the metallic aluminum that suffers an oxidation reaction:

Al (s) -> Al3 + (aq) + 3e–

and another is a reduction reaction involving copper;

2e– + Cu2 + (aq) -> Cu (s)

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How does plants react with no scratch?
Andreas93 [3]

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You see, plants need energy to grow and grow and grow. They use energy from sunlight to make a simple sugar, glucose. Whenever the plant needs energy, it can chomp a little glucose off of the starch.

without energy it wont get energy/food so it will eventually die  

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3 0
3 years ago
A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path
Igoryamba

Explanation:

  • For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

              R = 8.314 J/(K mol),          T = 298 K ,

          V_{2} = 8.34 L,    V_{1} = 2.67 L

Now, putting the given values into the above formula as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

   = -2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})

     = -2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494

    = -2818.68 J

Hence, work for path A is -2818.68 J.

  • For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure P_{external} = 1.00 atm.

So,              W = -P_{external} \times \Delta V

Now, putting the given values into the above formula as follows.

               W = -P_{external} \times \Delta V

                   = -1 atm \times (8.34 L - 2.67 L)  

                    = -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

       W = -\frac{101.33 J}{1 atm L} \times 5.67 atm L

            = -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

                        W = -574.54 J

Hence, work for path B is -574.54 J.

3 0
4 years ago
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