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timurjin [86]
3 years ago
12

A backyard swimming pool holds 150 cubic yards (yd3) of water. what is the weight of the water in pounds?

Physics
1 answer:
Crank3 years ago
5 0

solution:

1yd^3=(0.9144m)^3\\
therefore,\\
150yd^3=150yd^3\times\frac{(0.9144m)^3}{1yd^3}\\
=114.68m^3
convert m^3 to cm^3by the conversion factor as follow\\
1m^3=(100)cm^3\\
114.68m^3=114.68\times\frac{(100cm)3}{1m^3}\\
convert cm^3 to mL by the conversion factor as follow\\
1cm^3=1mL\\
114.68\times10^6cm^3=114.68\times10^6cm^3\times\frac{1mL}{1cm^3}\\
=114.68\times10^6mL\\
convert mL to g by the conversion factor as follow\\
1mL=1g\\therefore,\\
114.68\times10^6mL=114.68\times10^6mL\times\frac{1g}{1mL}\\
114.68\times10^6g\\
convert g to kg the conversion factor as follow\\
1g=10^-3kg\\
therefore,\\
114.68\times10^6=114.68\times10^6g\times\frac{10^-3kg}{1g}\\
=114.68\times10^3kg\\
convert kg to Ib by the conversion factor as follow\\
1kg=2.2Ib\\
therefore,\\
114.68\times10^3kg=114.68\times10^3kg\times\frac{2.2Ib}{1kg}\\
=2.52\times10^5Ib

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Answer:

Physical Properties of Sodium

Atomic number 11

Melting point 97.82°C (208.1°F)

Boiling point 881.4°C (1618°F)

Volume increase on melting 2.70%

Latent heat of fusion 27.0 cal/g

Lenntech Water treatment & purification

Toggle navigation

Home Periodic table Elements Sodium

Sodium - Na

Chemical properties of sodium - Health effects of sodium - Environmental effects of sodium

Atomic number

11

Atomic mass

22.98977 g.mol -1

Electronegativity according to Pauling

0.9

Density

0.97 g.cm -3 at 20 °C

Melting point

97.5 °C

Boiling point

883 °C

Vanderwaals radius

0.196 nm

Ionic radius

0.095 (+1) nm

Isotopes

3

Electronic shell

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Energy of first ionisation

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8 0
2 years ago
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3 years ago
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kolezko [41]

A. Wedge

and

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3 years ago
Two 100 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Bat -10 m/s when th
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3 0
3 years ago
Find the final temperature of 375 grams of tea (c = 4.184 J/g°C) if its initial temperature is 95°C just before it is placed in
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Answer:

the final temperature of the tea is 7.39⁰C.

Explanation:

Given;

mass of the tea, m = 375 g

specific heat capacity of the tea, C = 4.184 JJ/g°C

initial temperature of the tea, t₁ = 95°C

the final temperature of the tea, t₂ = ?

Energy lost by the refrigerator, Q = 137,460 J

The energy lost by the refrigerator is given by the following formula;

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-137,460 = 1569(t₂ - 95°C)

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Therefore, the final temperature of the tea is 7.39⁰C.

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