If iron has a density of 7.87g/cm³ and a mass of 3.729g, then the volume of iron is 0.474cm³
HOW TO CALCULATE VOLUME:
- The volume of a substance can be calculated by dividing the mass by its density. That is;
Volume (mL) = mass (g) ÷ density (g/mL)
- The density of iron is given as 7.87g/cm³ while its mass is 3.729g of iron. Hence, the volume can be calculated as follows:
Volume = 3.729 ÷ 7.87
Volume = 0.474cm³
Therefore, the volume of iron is 0.474cm³
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Answer:
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Explanation:
Answer:
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Explanation:
this is the answer
The SI base unit for length is meter.
In order to make smaller measurements, you can use the centi-, milli-, micro-, etc. prefixes.
When you want to reference larger measurements, you can use the kilo-, mega-, giga- and prefixes such as those.
<u>Answer:</u> The number of gold atoms per cubic centimeters in the given alloy is 
<u>Explanation:</u>
To calculate the number of gold atoms per cubic centimeters for te given silver-gold alloy, we use the equation:

where,
= number of gold atoms per cubic centimeters
= Avogadro's number = 
= Mass percent of gold in the alloy = 42 %
= Density of pure gold = 
= Density of pure silver = 
= molar mass of gold = 196.97 g/mol
Putting values in above equation, we get:

Hence, the number of gold atoms per cubic centimeters in the given alloy is 