1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
damaskus [11]
3 years ago
7

The intake stroke of a positive displacement compressor is most similar to that of a

Chemistry
1 answer:
umka21 [38]3 years ago
5 0

Answer:

D. two-stroke cylinder engine.

Explanation:

<u>Two-stroke cylinder engine:</u> In chemistry, the term "two-stroke cylinder engine" is described as one of the types of "internal combustion engine" that tends to complete a single "power cycle" possessing two different strokes of the "piston" during a particular "crankshaft revolution".

<u>It generally</u> consists of two different strokes that are being often known as power or exhaust and suction stroke.

<u>In the question above, the given statement signifies the "two-stroke cylinder engine". </u>

You might be interested in
What are the 5 areas of chemistry
shepuryov [24]
The 5 main branches of chemistry are physics, analytical, biochemistry, organic chemistry, and inorganic chemistry.
7 0
3 years ago
Please help me out i will give you brainlist. 0.500 is wrong
Alik [6]
<h3>Answer:  b) 0.250 mol</h3>

============================================

Work Shown:

Using the periodic table, we see that

  • 1 mole of carbon = 12 grams
  • 1 mole of oxygen = 16 grams

These are approximations and these values are often found underneath the atomic symbol. For example, the atomic weight listed under carbon is roughly 12.011 grams. I'm rounding to 2 sig figs in those numbers listed above.

So 1 mole of CO2 is approximately 12+2*16 = 44 grams. The 2 is there since we have 2 oxygens attached to the carbon atom.

-------------------

Since 1 mole of CO2 is 44 grams, we can use that to convert from grams to moles.

11.0 grams of CO2 = (11.0 grams)*(1 mol/44 g) = (11.0/44) mol = 0.250 mol of CO2

In short,

11.0 grams of CO2 = 0.250 mol of CO2

This is approximate.

We don't need to use any of the information in the table.

3 0
3 years ago
Read 2 more answers
if you placed the contents of a packet of powdered iced tea mix into a bottle of water and shook it, which of the following woul
victus00 [196]
If you placed the contents of a packet of powdered iced tea mix into a bottle of water and shook it, D. The solution would be homogenous if half of the powdered solute sat at the bottom of the bottle.
3 0
3 years ago
Read 2 more answers
What is the specific heat for the steel wire
olganol [36]
I think that the answer is 0.49
8 0
3 years ago
A sample of Ca(OH)2 is considered to be an Arrhenius base because it dissolves in water to yield
postnew [5]
The answer is C. Arrhenius bases increase the concentration of OH- in solution.
8 0
2 years ago
Read 2 more answers
Other questions:
  • The mole fraction of hcl in a solution prepared by dissolving 5.5 g of hcl in 200 g of c2h6o is ________. the density of the sol
    8·1 answer
  • When you balance a chemical reaction, you are making sure that the law of conservation of matter is obeyed?
    7·2 answers
  • For which of the following processes would there be a decrease in entropy?
    13·1 answer
  • What is a wind turbine often used for?
    13·2 answers
  • How many joules of energy are made by the loss of 100kg of mass using the formula E=mc2? please help!!!
    10·1 answer
  • From the information presented in the movie, what can you
    5·1 answer
  • In metallic bonds, electrons are referred to as electrons.​
    7·1 answer
  • The photo shows a clay pot being made on a spinning potter's wheel. What
    14·1 answer
  • How many moles of AgCl can form from 6.5 moles of Ag and abundant HCl in the equation: 2Ag + 2HCl -&gt; 2AgCl + H2
    12·1 answer
  • If an object does Not explode, catch fire, or dissolve, how would you describe this object? (Three answers)​
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!