If a question cannot be tested or observed it cannot be answered by science? True or false??

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

In-s [12.5K]

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}$O during the }\boxed{\textbf{exhaust stroke}}$

7 0

3 years ago

What is formed when 2 or more atoms are held together by covalent bonds?

Ksenya-84 [330]

The group of atoms held together by covalent bonds is called a molecule

7 0

4 years ago

g Consider an ideal atomic gas in a cylinder. The upper part of the cylinder is a moveable piston of negligible weight. The heig

kumpel [21]

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

A cylindrical weight with a mass (m) of 3 kg is dropped, that is, its initial velocity (u) is 0 m/s and travels 10 m (s). Assuming the acceleration (a) is that of gravity (9.8 m/s²). We can calculate the velocity (v) of the weight in the instant prior to the collision with the piston using the following kinematic equation.

$v^{2} = u^{2} + 2as = 2 (9.8 m/s^{2} ) (10m) \\\\v = 14 m/s$

The object with a mass of 3 kg collides with the piston at 14 m/s, The kinetic energy (K) of the object at that moment is:

$K = \frac{1}{2} m v^{2} = \frac{1}{2} (3kg) (14m/s)^{2} = 294 J$

The kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Thus, Q = 294 J.

Given all the process is at 250 K (T), we can calculate the change of entropy of the gas using the following expression.

$\Delta S_{gas} = \frac{Q}{T} = \frac{294 J}{250K} = 1.18 J/K$

The change in the entropy of the environment, has the same value but opposite sign than the change in the entropy of the gas. Thus, $\Delta S_{env} = -1.18 J/K$

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

Learn more: brainly.com/question/22655760

6 0

2 years ago

For the balanced equation shown below, if the reaction of 0.112 grams of

Alekssandra [29.7K]

Answer:

The answer is 74.5%.

Explanation:

As we know that % yield= $\frac{actual yield}{theoretical yield}$ x 100%.

Therefore,

Step 1 Calculate Theoretical yield:

0.112 $H_{2}$ x $\frac{1 mol H_{2} }{2.016 g of H{2} }$ x $\frac{4 mol H_{2}O }{4 mol H_{2} }$ x $\frac{18.02 g H_{2}O }{1 mol H_{2}O}$ = 1.001 g $H_{2}O$