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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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Molality of the solution = 0.7294 M

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Given:

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3 years ago

Unit: Stoichiometry

Reika [66]

Answer:

1. 2.41 × 1023 formula units

2. 122 L

3. 7.81 L

Explanation:

1. Equation of the reaction: 2 Na(NO3) + Ca(CO3) ---> Na2(CO3) + Ca(NO3)2

Mole ratio of NaNO3 to CaCO3 = 2 : 1

Moles of CaCO3 = mass/molar mass

Mass of CaCO3 = 20 g; molar mass of CaCO3 = 100 g

Moles of CaCO3 = 20 g/100 g/mol = 0.2 moles

Moles of NaNO3 = 2 × 0.2 moles = 0.4 moles

1 Mole of NaNO3 = 6.02 × 10²³ formula units

0.4 moles of NaNO3 = 0.4 × 6.02 × 10²³ = 2.41 × 1023 formula units

2. Equation of reaction : 2 H2O ----> 2 H2 + O2

Mole ratio of oxygen to water = 1 : 2

At STP contains 6.02 × 10²³ molecules = 1 mole of water

6.58 × 10²⁴ molecules = 6.58 × 10²⁴ molecules × 1 mole of water/ 6.02 × 10²³ molecules = 10.93 moles of water

Moles of oxygen gas produced = 10.93÷2 = 5.465 moles of oxygen gas

At STP, 1 mole of oxygen gas = 22.4 L

5.465 moles of oxygen gas = 5.465 moles × 22.4 L/1 mole = 122 L

3.Equation of reaction: 6 K + N2 ----> 2 K3N

Mole ratio of Nitrogen gas and potassium = 6 : 1

Moles potassium = mass/ molar mass

Mass of potassium = 90.0 g, molar mass of potassium = 39.0 g/mol

Moles of potassium = 90.0 g / 39.0 g/mol = 2.3077moles

Moles of Nitrogen gas = 2.3077 moles / 6 = 0.3846 moles

At STP, 1 mole of nitrogen gas = 22.4 L

0.3486 moles of oxygen gas = 0.3486 moles × 22.4 L/1 mole = 7.81 L

7 0

3 years ago