Average density of a copper-coated iron BB is 8.00g/mL. If the density for copper is 8.92 g/mL, and the density for iron is 7.43g/mL, what is the percentages of copper and iron in a BB?
Notice that mass is density*volume.
total mass=massiron*densityiron*volume + masscopper*densitycopper*volume
Now, divide by total mass
1= fractioniron*densityiron*volume/mass + fractionCu*densityCu*volume/mass.
now note that volume/mass=1/avgdensity
and that fractionCu=1-fractioniron
That should be enough for you to grunt through the algebra.
I am confused on what the fraction is?? and what is my volume? could you please clear this up?
To find the percentages of copper and iron in a copper-coated iron BB, we need to first understand the concept of density and its relationship to mass and volume.
Density is defined as the mass of a substance per unit volume. The formula for density is:
Density = mass / volume
In this case, we are given the density of the copper-coated iron BB (8.00 g/mL), as well as the individual densities of copper (8.92 g/mL) and iron (7.43 g/mL).
Now, let's denote the percentage of copper in the BB as X% and the percentage of iron as Y%. Since the BB is a combination of both copper and iron, the sum of X% and Y% should add up to 100%.
We also know that the density of the copper-coated iron BB is the weighted average of the densities of copper and iron, based on the percentage of each element present. Mathematically, this can be expressed as:
Density of BB = (Percentage of copper * Density of copper) + (Percentage of iron * Density of iron)
Substituting the given values, we get:
8.00 g/mL = (X% * 8.92 g/mL) + (Y% * 7.43 g/mL)
Now, we have two unknowns (X% and Y%) and one equation. To solve for X% and Y%, we need another equation.
Since X% and Y% represent the percentages of copper and iron respectively, we can write a second equation:
X% + Y% = 100%
This equation ensures that the sum of the percentages equals 100%.
Now, we have a system of two equations:
8.00 = (X% * 8.92) + (Y% * 7.43) (Equation 1)
X% + Y% = 100% (Equation 2)
To solve this system of equations, we can use various methods such as substitution or elimination. Let's use the substitution method in this case.
Rearranging Equation 2, we get:
X% = 100% - Y%
Substituting this value of X% in Equation 1, we get:
8.00 = ((100% - Y%) * 8.92) + (Y% * 7.43)
Now, we can solve this equation for Y%:
8.00 = (892 - 8.92Y) + 7.43Y
8.00 = 892 + (-8.92Y + 7.43Y)
8.00 = 892 - 1.49Y
-1.49Y = 892 - 8.00
-1.49Y = -116
Y ≈ -116 / -1.49 ≈ 77.85%
Now that we have the value of Y%, which represents the percentage of iron in the BB, we can substitute it back into Equation 2 to find X%:
X% = 100% - Y%
X% = 100% - 77.85%
X% ≈ 22.15%
Therefore, the percentage of copper in the copper-coated iron BB is approximately 22.15%, and the percentage of iron is approximately 77.85%.