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Pani-rosa [81]
2 years ago
12

Water flows over Niagara Falss at the average rate of 2,400,000 kg/s, and the average height of the falls is about 50 m. Knowing

that the graviatational potential energy of falling water per second = mass (kg) x height (m) x gravity (9.8 m/s2), what is the power of Niagara Falls? How many 15 W LED light bulbs could it power?
Chemistry
1 answer:
liberstina [14]2 years ago
4 0

Answer:

1. 176 × 10^12 W ; 78400000000

Explanation:

Given the following :

Fall rate = 2,400,000kg/s

Average height of fall = 50m

Gravitational Potential of falling water = mgh = mass × acceleration due to gravity × height =

How many 15 W LED light bulbs could it power?

Recall : power = workdone / time

Workdone = gravitational potential energy

Mass of water = density * volume

Density of water = 1 * 10^3kg/m^3

Rate of fow = volume / time = 2400000

Hence,

Power = 1000 * 2,400,000 * 9.8 * 50

Power = 1176000000000

Power = 1. 176 × 10^12 W

How many 15 W LED light bulbs could it power?

1176000000000 / 15 = 78400000000

= 78400000000 15 W bulbs

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495 cm3 of oxygen gas and 877 cm3 of nitrogen gas, both at 25.0 C and 114.7 kpa, are injected into an evacuated 536 cm3 flask. F
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Answer:

<u><em>Total pressure of the flask is 2.8999 atm.</em></u>

Explanation:

Given data:

Volume of oxygen (O2) gas= 495 cm3

                                              = 0.495 L (1 cm³ = 1 mL = 0.001 L)                                            

Volume of nitrogen (N2) gas =  877 cm3

                                               = 0.877 L (1 cm³ = 1 mL = 0.001 L)

volume of falsk = 536 cm3

                         = 0.536 L (1 cm³ = 1 mL = 0.001 L)

Temperature =  25 °C

T = (25°C + 273.15) K

    = 298.15 K

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               = 114.700 Pa

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                        = 1.132 atm

Formula:

PV=nRT  <em>(ideal gas equation)</em>

P = pressure

V = volume

R (gas constnt)=  0.0821 L.atm/K.mol

T = temperature

n = number of moles for both gases

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Firstly we will find the number of moles for oxygen and nitrogen gas.

<u>For Oxygen:</u>

n = PV / RT

n = 1.132 atm × 0.495 L / 0.0821 L.atm/K.mol × 298.15 K

  = 0.560 / 24.47

  = 0.0229 moles

<u>For Nitrogen:</u>

n = PV / RT

n = 1.132 atm × 0.877 / 0.0821 L.atm/K.mol × 298.15 K

n = 0.992 / 24.47

  = 0.0406

Total moles = moles for oxygen gas + moles for nitrogen gas

  = 0.0229 moles + 0.0406 moles

n  = 0.0635 moles

Now put the values in formula

PV=nRT

P = nRT / V

P = 0.0635 × 0.0821 L.atm/K.mol × 298.15 K  /  0.536 L

P = 1.554 / 0.536

<u><em>P = 2.8999 atm</em></u>

Total pressure in the flask is  2.8999 atm, while assuming the temperature constant.

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