How do I convert kg/m3 to g/cm3 and understand the logic behind it?

In summary, the person is trying to figure out how to convert 3200 kg/m3 to g/cm3. They explain that dimensions are matched so the answer is correct, but they don't understand why the coefficiant (value in front) is in either spot. They ask for help understanding why this is the case and are told that it is all about "asking yourself".
  • #1
Nellen2222
55
0

Homework Statement




How do I convert kg/m3 to g/cm3?.

3200kg/m^3 ---> 3.2g/cm^3


Homework Equations




The question was 3200kg/m3 = 3.2g/cm3. I don't understand the steps involved?

The Attempt at a Solution



I know how to do dimensional analysis, but I just don't really understand the logic in this. I'll go step by step and please explain this to me it is crucial!


1) 3200kg/m^3 ----> g/cm^3

2) 3200kg/m^3 * 1000kg/1g*1m^3/1*10^-6cm^3 = 3.2kg/cm^3

I just don't understand why I divide g/m^3 by a million to reach g/cm^3. Isnt one cm^3 a smaller value(or area) than a m^3? Shouldn't I be multiplying by a million?

For instance: if I don't have a density and simply convert 100m^3 to 100cm^3 the answer is 100 000 000 cm^3. This makes sense because a metre is larger than a CM thus the value will be bigger .

So, Why is this not the case for when I have a density? Why do I have to divide by a million in order to go from m^3 to cm^3 in that case and ultimately reach a smaller value.

Please help me understand where my logic goes wrong.
 
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  • #2
3200 kg m-3 = 3200 kg m-3 X 1000 g kg-1/ 100000 cm3 m-3
 
  • #3
? what
 
  • #4
Convert kg into g. (multiplying 1000 g per kg)

Convert cubic metre into cubic centimetre (dividing 1000000 cc per cubic metre)

Dimensions are matched, so we get the right answer.
 
  • #5
The dimensions can match but the coefficiant(value in front) can be in either spot.. so technically can still have the wrong answer..
 
  • #6
Man, they are correct. If you are adamant to make mistake, even God can't help it then. Using dimensions in coefficients helps us to determine if we are going on the right path.

Ask yourself - How many grams in a kilogram? How many cc in a litre -> how many litres in a cubic metre?
 
  • #7
yeah- its all about 'ask yourself' - I just don't get why I am dividing by a million if a m^3 is bigger than a metre^3..
 
  • #8
? What are you saying!? Isn't metre is shorted as m?
 

FAQ: How do I convert kg/m3 to g/cm3 and understand the logic behind it?

1. What is dimensional analysis and why is it important in science?

Dimensional analysis is a mathematical technique used to check the consistency of units in a problem. It involves converting values from one unit to another using conversion factors. It is important in science because it helps ensure that calculations and measurements are accurate and consistent, and can help identify errors in equations or calculations.

2. How is dimensional analysis used in chemistry?

Dimensional analysis is commonly used in chemistry to convert between different units of measurement, such as converting from grams to moles or from liters to milliliters. It also helps in performing unit conversions for different chemical equations and stoichiometric calculations.

3. Can dimensional analysis be used in physics?

Yes, dimensional analysis is widely used in physics to check the correctness of equations and to convert between different units of measurement. It is particularly useful in solving problems related to motion, force, energy, and other physical quantities.

4. What are some common conversion factors used in dimensional analysis?

Common conversion factors used in dimensional analysis include the metric system prefixes (kilo, centi, milli, etc.), Avogadro's number, the ideal gas constant, and other fundamental physical constants. These conversion factors allow for easy conversion between different units of measurement in a given system.

5. How can dimensional analysis help in problem-solving?

Dimensional analysis can help in problem-solving by providing a systematic approach to converting between units and checking the accuracy of calculations. It can also help identify errors and inconsistencies in equations, making it a useful tool for troubleshooting and problem-solving in various scientific fields.

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