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Anni [7]
3 years ago
15

An automobile is driving uphill. Which form of energy is not involved in this process?

Chemistry
1 answer:
Lapatulllka [165]3 years ago
7 0
The answer is D) Electromagnetic energy. This is not a type of energy that occurs when a vehicle is moving.
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Predictable moderate winds between 5 & 25 degrees North or South latitude are called?
Grace [21]

Answer:

احبـــــــــــــــــــك

لول ملل اقولج انا اسفه

6 0
2 years ago
Are weathering and erosion two different names for the same process? Explain why or why not.PLEASE HELP!
Katyanochek1 [597]

Answer:

While weathering and erosion are similar processes, they are not synonymous. Weathering involves the breakdown of rocks and minerals on Earth, whereas erosion involves the removal of soil and rock materials.

6 0
3 years ago
A certain substance melts at a temperature of . But if a sample of is prepared with of urea dissolved in it, the sample is found
pshichka [43]

Answer:

2.2 °C/m

Explanation:

It seems the question is incomplete. However, this problem has been found in a web search, with values as follow:

" A certain substance X melts at a temperature of -9.9 °C. But if a 350 g sample of X is prepared with 31.8 g of urea (CH₄N₂O) dissolved in it, the sample is found to have a melting point of -13.2°C instead. Calculate the molal freezing point depression constant of X. Round your answer to 2 significant digits. "

So we use the formula for <em>freezing point depression</em>:

  • ΔTf = Kf * m

In this case, ΔTf = 13.2 - 9.9 = 3.3°C

m is the molality (moles solute/kg solvent)

  • 350 g X ⇒ 350/1000 = 0.35 kg X
  • 31.8 g Urea ÷ 60 g/mol = 0.53 mol Urea

Molality = 0.53 / 0.35 = 1.51 m

So now we have all the required data to <u>solve for Kf</u>:

  • ΔTf = Kf * m
  • 3.3 °C = Kf * 1.51 m
  • Kf = 2.2 °C/m
5 0
2 years ago
What are the molality and mole fraction of solute in a 22.3 percent by mass aqueous solution of formic acid (HCOOH)?
Vanyuwa [196]

Answer:

Mole fraction for solute = 0.1, or 10%

Molality = 6.24 mol/kg

Explanation:

22.3% by mass → In 100 g of solution, we have  22.3 g of HCOOH

Mass of solution = 100 g

Mass of solute = 22.3 g

Mass of solvent = 100 g - 22.3g = 77.7 g

Let's convert the mass to moles

22.3 g . 1mol/ 46 g = 0.485 moles

77.7 g. 1mol / 18 g = 4.32 moles

Total moles = 4.32 moles + 0.485 moles = 4.805 moles

Xm for solute = 0.485 / 4.805 = 0.100 → 10%

Molality → mol/ kg → we convert the mass of solvent to kg

77.7 g.  1 kg / 1000g = 0.0777 kg

0.485 mol / 0.0777 kg = 6.24 m

6 0
3 years ago
Read 2 more answers
Question 6
I am Lyosha [343]

Answer:

A. is the correct point.

Explanation:

This is true because no matter how many mL of water is added, the solution only gets more height; the concentration in everything else stays the same, and water doesn't have any concentration. Very confusing, I know. Good luck!

4 0
2 years ago
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