Chemistry help?

a water bed filled with water has the dimensions 8.0 ft by 7.0 ft by 0.75 ft. Taking the density of water to be 1.00g/ cm^3, how many kilograms of water are required to fill the water bed?

Answer:
mass/volume = density
mass = volume x density
Density = 1 g/cm^3 = 0.001kg/cm^3

volume = 8x7x0.75
=42 cu.ft
convert cubic.ft to cu.cm (consider 1 ft = 30cm)
volume = 1134000 cubic.cm or 1.1 x 10^6 cu.cm

mass = (1134000)(0.001)
mass = 1134 kg or 1.1 x 10^3 kg
Volume = 8x7x0.75 cubic feet
Mass (m) = Denisty (d) x Voume (v)
m=1g/cm 3x volume converted into cubic centimetre
1189.31 Kg

This is from

8ft x 7ft x 0.75 ft = 42 cubic feet.

2.54 cm/inch
and 30.48 cm/foot

28316.846592 cubic cm/cubic foot.

42 cubic feet = 1189307.556864 cubic cm

This is then equal to 1189307.556864 grams

Simply dividing by 1000g/Kg gives us the answer of 1189.31 Kg.




Be careful about some of these other answers, they introduce too much error through early rounding. I believe that mine has the most correct answer since I only round at the end of the problem.
ok,. this problem is easy, u need found the mass isnt ture?
so what is the information that we have?

density = 1.00g/cmª = 1000kg/m3 because u need the answer in kg. i change gm to kg
´then the dimension are 8X7x0.75 ft change to meters please. (1 ft =0.3048mts) then the volumen =12.8016 ( volumen is LargeXHighXwide)
if density =m/vol , mass =density*volument
masa = densityXvolumen then 1000kg/m3 * 12.8016=12,8016 kg.
Well basically your waterbed is 8ftx7.0ftx0.75ft= 42ft3

Now use dimentional analysis:

42ft3x1728in3x16.4cm3= 1.2E6cm3
.....1ft3...1in3

1.2E6cm3x1.00g.x.1kg...= 1200kg of water
.....1cm3..1000g
Volume = 8 x 7 x 0.75 = 42 ft³
1 ft³ of water = 62.4 lbs = 62.4 x 42 = 2,621 lbs.
2,621lbs ÷ 2.2 lb/kg = 1,191.3kg of water.

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