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spayn [35]
3 years ago
12

If a water fountain has an average flow rate of 0.64 gallons per minute, how many

Chemistry
1 answer:
nekit [7.7K]3 years ago
3 0

The time taken for the water fountain to dispense 842.75 mL of water is 20.625 seconds

<h3 /><h3>What is time?</h3>

Time can be defined as the measure of past, present or future events.

To calculate the time taken to dispense 842.75 mL of water, we use the formula below.

Formula:

  • F = V/t................ Equation 1

Where:

  • F  = Average flow rate of the fountain
  • V = Volume of the water
  • t = time.

Make t the subject of the equation.

  • t = V/F................ Equation 2

From the question,

Given:

  • F = 0.64 gallons per minutes
  • V = 842.75 mL = (842.75/1000)/(3.78541) gal = 0.22 gal.

Substitute these values into equation 2

  • t = 0.22/0.64
  • t =  0.34375 minutes.
  • t = (0.34375×60) =  20.625 seconds.

Hence, The time taken for the water fountain to dispense 842.75 mL of water, is 20.625 seconds.

Learn more about time here: brainly.com/question/13893070

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A 3.8-mol sample of KClO3 was decomposed according to the equation. How many moles of O2 are formed assuming 100% yield?
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Answer:

5.7 moles of O2

Explanation:

We'll begin by writing the balanced decomposition equation for the reaction. This is illustrated below:

2KClO3 —> 2KCl + 3O2

From the balanced equation above,

2 moles of KClO3 decomposed to produce 3 moles of O2.

Next, we shall determine the number of mole of O2 produced by the reaction of 3.8 moles of KClO3.

Since 100% yield of O2 is obtained, it means that both the actual yield and theoretical yield of O2 are the same. Thus, we can obtain the number of mole of O2 produced as follow:

From the balanced equation above,

2 moles of KClO3 decomposed to produce 3 moles of O2.

Therefore, 3.8 moles of KClO3 will decompose to produce = (3.8 × 3)/2 = 5.7 moles of O2.

Thus, 5.7 moles of O2 were obtained from the reaction.

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How does heat always flow or transfer between objects or places
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4 years ago
A certain organ pipe, open at both ends, produces a fundamental frequency of 272 HzHz in air. Part A If the pipe is filled with
vladimir2022 [97]

Answer:

fundamental frequency in helium = 729.8 Hz

Explanation:

Fundamental frequency of an ope tube/pipe = v/2L

where v is velocity of sound in air = 340 m/s; λ is wave length of wave = 2L ; L  is length of  the pipe

To find the length of the pipe,

frequency  = velocity of sound / 2L

272 = 340 / 2 L

L = 0.625 m

If the pipe is filled with helium at the same temperature, the velocity of sound will change as well as the frequency of note produced since velocity is directly proportional to frequency of sound.

Also, the velocity of sound is inversely proportional to  square root of molar mass of gas; v ∝ 1/√m

v₁/v₂ = √m₂/m₁

v₁ = velocity of sound in air, v₂ = velocity of sound in helium, m₁ = molar mass of air, m₂ = molar mass of helium

340 / v = √4 / 28.8

v₂ = 340 / 0. 3727

v₂  =  912.26  m /s  

fundamental frequency in helium  = v₂ / 2L

fundamental frequency in helium = 912.26 / (2 x 0.625)

fundamental frequency in helium = 729.8 Hz

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Chemical formula Be+2 + N-3
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