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You would have to know the density also.

Mass = volume X density

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Q: What is the mass if the volume is 25cm cubed?
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A brick has a mass of 100g and a volume of 25cm cubed. what is the density of the brick?

Density is defined as mass divided by volume, therefore: 100g/25cm3 = 4 g/cm3


What is the formula for density of a object using mass and volume cubed?

Density = Mass/VolumeVolume is not cubed, although the units in which volume is expressedmay be cubed units.


What is the volume of a 25Cm metal cube?

A cube which has edges 25cm long has a volume of 25 cubed cubic centimetres = 25 x 25 x 25= 15625 cubic centimetres (cc) =15.625 litres


How do you get mass with volume and density?

density = mass/volume volume = mass/density mass = density x volume Ex: V=40m cubed M=300g D=a M/V 300/40 = 7.5g _______ m cubed or D= 7.5 grams per meter cubed


What is 25cm cubed?

15625 cubic cm


What is the mass of an object its density is 5 grams centimeters cubed and the volume is 10 centimeters cubed grams to centimeters cubed converter grams to centimeters cubed conversion?

mass = density x volume Mass = 5 grams per centimeter cubed x 10 centimeters cubed = 50 grams


What is The mass of 5.42 cm cubed of platinum if the density is 22.5 cm cubed?

The density must be 22.5 GRAMS per cm cubed, not 22.5 cm cubed. Then, Density = Mass/Volume implies Mass = Density*Volume = 22.5 * 5.42 = 121.95 grams.


What is the mass of gold if it has a volume of 10cm cubed and its density was 19.3g cm cubed?

193g


Measurement of mass in a volume of a substance?

units cubed


What is the density of an object with a mass of 50 grams and a volume of 5cm cubed?

Just divide the mass by the volume.


If density of an object is 3g per centimeter cubed what is the mass?

You need to know the volume: Mass = volume X density


What is the density of an object having a mass of 0.8g and a volume of 25cm?

Not enough information. To get the density, you need to divide the mass by the volume, but there is no way of knowing the volume of an object just by having one of its lengths.