The density of wind-packed snow is estimated to be 0.35 g/cm^3. A flat roof that is 35 by 43 feet has 28 inches of snow on it. How many pounds of snow are on the roof?

There are 76728.2 pounds of snow on the roof

Explanation:

To determine how many pounds of snow are on the roof,

First we will determine the mass of the snow in grams.

From the question,

Density of snow = 0.35 g/cm³

From

Density = Mass / Volume

To determine the mass, we will first find the volume.

From the question, a flat roof that is 35 by 43 feet has 28 inches of snow on it, that is, the height of the snow on the roof is 28 inches.

Area of the roof = 35 ft × 43 ft = 1505 ft²

Volume of the snow = Area of the roof × Height of the snow

Height of the snow = 28 inches

12 inches = 1 foot

∴ 28 inches = 28/12 feet = 2.3333 ft

Hence,

Volume of the snow = 1505 ft² × 2.3333 ft = 3511.62 ft³

Volume of the snow = 3511.62 ft³

This is the volume of the snow in cubic feet

Now, we will convert cubic feet (ft³) to cubic centimeter (cm³)

1 ft = 30.48 cm

and 1 ft³ = 30.48³ cm³ = 28316.84659 cm³.

If 1 ft³ =  28316.84659 cm³

∴ 3511.62 ft³ will be 3511.62 × 28316.84659 cm³ = 99438004.83 cm³

Hence, Volume of snow = 99438004.83 cm³

Now, we can determine mass of snow in grams

From

Density = Mass / Volume

Mass = Density × Volume

Mass = 0.35 g/cm³ × 99438004.83 cm³

Mass = 34803301.69 g

Now, we will convert the mass from grams to pounds

1 pound = 453.592 grams

If 453.592 grams = 1 pound

Then, 34803301.69 g will be 34803301.69 / 453.592 pounds = 76728.2 pounds

Hence, there are 76728.2 pounds of snow on the roof.

i believe it is c.) 1575 j.

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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