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3 years ago

In which step of a four-stroke engine cycle does the car release CO2, H2O, and CO?

Chemistry

1 answer:

In-s [12.5K]3 years ago

7 0

Answer:

$\boxed{\text{exhaust stroke}}$

Explanation:

The four strokes are

Intake: The mixture of air and fuel enters the cylinder.
Compression: The mixture is compressed by a factor of about 11.
Ignition: A spark ignites the fuel, which burns and produces CO₂, CO, and H₂O.
Exhaust: The combustion products are expelled from the cylinder.

$\text{The cylinder releases CO$_{2}$, CO, and H$_{2}$O during the }\boxed{\textbf{exhaust stroke}}$

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The very common mineral shown in this photograph is commonly a pink- to cream-colored mineral with wavy, light-colored lines. it

musickatia [10]

The very common mineral shown in the figure that is referred in this problem that is commonly a pink- to cream-colored mineral with wavy, light-colored lines and does not effervesce would be feldspar. It make up about 41 percent weight of the Earth's crust. It is a group of rocks that contains tectosilicate compounds.

8 0

3 years ago

There are several aromatic compounds with the formula C8H9Cl. Draw those that have a trisubstituted ring where the methyl groups

Kryger [21]

Answer:

$3-Chloro-1,2-dimethylbenzene$

$4-Chloro-1,2-dimethylbenzene$

Explanation:

Such types of compounds in which conjugated planer ring system and delocalized pi electrons are present are called aromatic compounds such as Toluene, Benzene and some other. Generally, these compounds follow Huckle's rule. The trisubstituted ring means a compound in which three hydrogen atoms are replaced by three other groups. In the given compound two hydrogen atoms are replaced by two methyl groups and one hydrogen atom is replaced by one Chlorine atom. From the given compound $3-Chloro-1,2-dimethylbenzene$ , and $4-Chloro-1,2-dimethylbenzene$ can be drawn.

7 0

3 years ago

At 400 K oxalic acid decomposes according to the reaction:H2C2O4(g)→CO2(g)+HCOOH(g)In three separate experiments, the intial pre

padilas [110]

Answer:

v = 2,66x10⁻⁵ P[H₂C₂O₄]

Explanation:

For the reaction:

H₂C₂O₄(g) → CO₂(g) + HCOOH(g)

At t = 0, the initial pressure is just of H₂C₂O₄(g). At t= 20000 s, pressures will be:

H₂C₂O₄(g) = P₀ - x

CO₂(g) = x

HCOOH(g) = x

P at t=20000 is:

P₀ - x + x + x = P₀+x. That means P at t=20000s - P₀ = x

For 1st point:

x = 92,8-65,8 = 27

Pressure of H₂C₂O₄(g) at t=20000s: 65,8-27 = 38,8

2nd point:

x = 130-92,1 = 37,9

H₂C₂O₄(g): 92,1 - 37,9 = 54,2

3rd point:

x = 157-111 = 46

H₂C₂O₄(g): 111-46 = 65

Now, as the rate law is :

v = k P[H₂C₂O₄]

Based on integrated rate law, k is:

(- ln P[H₂C₂O₄] + ln P[H₂C₂O₄]₀) / t = k

1st point:

k = 2,64x10⁻⁵

2nd point:

k = 2,65x10⁻⁵

3rd point:

k = 2,68x10⁻⁵

The averrage of this values is:

k = 2,66x10⁻⁵

That means law is:

v = 2,66x10⁻⁵ P[H₂C₂O₄]

I hope it helps!

4 0

3 years ago

Does it require more energy to "vaporize" water at the boiling point or to melt water at the melting point? Explain.

sveticcg [70]

Answer:

yes I think so because more heat evaporates water

7 0

3 years ago

Calculate the mass in grams for 2.28 moles of N2.

borishaifa [10]

Answer: 63.9

Explanation:

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2 years ago