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prisoha [69]
3 years ago
15

A factory at a low elevation ships bags of popcorn to cities at high elevations. Why does the factory manager need to monitor th

e amount of air in each bag
Chemistry
2 answers:
motikmotik3 years ago
8 0

Explanation:

Since factory is at low elevation, therefore, there will be more pressure. Similarly at high elevations there will be low pressure.

So, when bags of popcorn move from low elevation to high elevation the factory manager need to monitor the amount of air in each bag because if there are too many molecules in the bag then it will pop as it goes from high to low pressure areas.



german3 years ago
7 0

A factory at a low elevation ships bags of popcorn to cities at high elevations. The factory manager need to monitor the amount of air in each bag because when the bags of popcorn are delivered into high elevation cities, it will shrink due to the high pressure exerted by the surroundings towards the bag. As the elevation increases, the pressure surrounding it also increases.

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Y_Kistochka [10]

Answer:

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Explanation:

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5 0
3 years ago
Read 2 more answers
How many moles are in 5.6x1025 molecules of calcium oxide?
Vlad1618 [11]

Answer:

<h2>93.02 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{5.6 \times  {10}^{25} }{6.02  \times  {10}^{23} }  \\  = 93.023...

We have the final answer as

<h3>93.02 moles</h3>

Hope this helps you

8 0
3 years ago
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5 0
3 years ago
If a car is traveling 100 km/h and comes to a stop in 3 minutes,what is acceleration of a passenger who is using vehicle restrai
Rama09 [41]

We are given –

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  • Initial velocity of car is, u= 100 km/hr
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Here–

\qquad\pink{\bf \longrightarrow  Initial\:  velocity =  100 \:km/hr}

\qquad\sf \longrightarrow  Initial\:  velocity = \dfrac{ 100 \times 1000}{3600} \:m/s

\qquad\pink{ \bf \longrightarrow  Initial\:  velocity = 27.78\: m/s}

Now –

\qquad____________________________

\qquad\purple{\bf \longrightarrow  Acceleration  = \dfrac{Final\: Velocity -Initial  \:Velocity }{Time}}

\qquad\purple{\bf \longrightarrow  Acceleration  = \dfrac{v -u}{t}}

\qquad\sf \longrightarrow  Acceleration = \dfrac{(0- 27.78)}{1800}

\qquad\sf \longrightarrow  Acceleration =\cancel{ \dfrac{- 27.78}{1800}}

\qquad\purple{\bf \longrightarrow  Acceleration = -0.15 \: m/s^2}

\qquad_______________________________

6 0
2 years ago
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pychu [463]

Answer:

Explanation:

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6 0
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