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lyudmila [28]
3 years ago
14

Which best describes the heat that you can feel on the engine of a running car?

Chemistry
2 answers:
IgorC [24]3 years ago
8 0
The heat is unusable  and  is being lost  to the surrounding -- assuming that the engine has reached operating temperature   and is not increasing or decreasing in temperature. 
butalik [34]3 years ago
7 0

Answer is: The heat is unusable and lost to the surroundings.

Gasoline is a mixture  of many different hydrocarbons: alkanes (paraffins), cycloalkanes  and alkenes (olefins).

Balanced chemical reaction of gasoline combustion:

C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O + energy.

One part of energy produced in combustion of gasoline is used for car ungine (kinetic energy) and other part of that energy (in form of heat) is lost to the surronding and engine does not use it.

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Which best describes how sediment forms??
Andreas93 [3]

Answer:

a.) the breakdown into small particles of rock and stuff

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3 years ago
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Brainliest to whoever gets it correct.
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I WILL MARK BRAINLISET FOR WHOEVER GETS THIS RIGHT! A pot is placed on a gas flame, and the water inside the pot begins to boil.
NNADVOKAT [17]

Answer:

B.     Warm water rises within the pot.?

Explanation

<em>There wasn't enough information given for me to safely determine the correct answer.</em>

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3 years ago
The decomposition of N2O to N2 and O2 is a first-order reaction. At 730°C, the rate constant of the reaction is 1.94 × 10-4 min-
grin007 [14]

Answer:

Total pressure 5.875 atm

Explanation:

The equation for above decomposition is

2N_2O \rightarrow 2N_2 + O_2

rate constant k =  1.94\times 10^{-4} min^{-1}

Half life t_{1/2} = \frac{0.693}{k} = 3572 min

Initial pressure N_2 O = 4.70 atm

Pressure after 3572 min = P

According to first order kinematics

k = \frac{1}{t} ln\frac{4.70}{P}

1.94\times 10^{-4} = \frac{1}{3572} \frac{4.70}{P}

solving for P we get

P = 2.35 atm

2N_2O \rightarrow 2N_2 + O_2

initial           4.70                         0             0

change        -2x                          +2x           +x

final             4.70 -2x                     2x           x

pressure ofO_2 after first half life  = 2.35 = 4.70 - 2x

                                                          x = 1.175

pressure of N_2 after first half life  =  2x = 2(1.175) = 2.35 ATM

Total pressure  = 2.35 + 2.35 + 1.175

                          = 5.875 atm

5 0
3 years ago
Read 2 more answers
How many moles of Mg3(PO4)2 are in 350.00 grams of Mg3(PO4)2?
dusya [7]

Answer:

1.3 moles/ 1.33150727 moles

Explanation:

350g x 1 mol/262.86g = 1.3 moles

6 0
3 years ago
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