The track along which electricity flows is called a:<br><br> fuse<br> circuit<br> switch

Over the course of many years, a rural farm town becomes more urban. Which of the following will most likely happen in the town

MrRa [10]

Answer:

Explanation:

Rivers may be contaminated by sewage.

I do hope I helped you! :)

6 0

3 years ago

What does the term "compound" mean? two or more substances dispersed in one another but each retaining their own identity. two o

GarryVolchara [31]

I believe it's the second option. 2 or more elements joined together such that the elements have lost their individual identity in favour of a new set of properties.

4 0

3 years ago

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5. Why doesn't the air at a frontal boundary mix?

Vsevolod [243]

Answer:

At a front, the two air masses have different densities, based on temperature, and do not easily mix. One air mass is lifted above the other, creating a low pressure zone.

Explanation:

Hope this helps!

4 0

3 years ago

Use the table to answer the question.

Alex17521 [72]

Answer:

High levels of moisture are contained within an air mass.

Tell me if I'm correct, plz!

Explanation:

6 0

3 years ago

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What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91

Vikentia [17]

Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

$1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L$

Temperature = $91^{o}C = (91 + 273) K = 364 K$

As molar mass of $O_2$ is 32 g/mol. Hence, the number of moles of $O_2$ are calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol$

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

$PV = nRT\\P \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm$

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

4 0

3 years ago

The track along which electricity flows is called a: fuse circuit switch