Processing 0.25% Silver Ore to Obtain 2 Pounds

[blue4882]

How many cubic centimeters of ore 0.25% by mass silver must be processed to obtain 2.0 pounds of silver? The density of the ore is 7.0 g/cm cubed. cm cubed; 1.00 pound=454 g. Assume that the process is 100% efficient.

I get the answer of 520 but the answer is 52000 and i don't understand why I am so off. Can someone help me please i would really appreciate it.


Thank you

 

[mrjeffy321]

This is just a guess, but since your answer is off by a factor of 100, I bet the problem came when you did not divide .25% by 100 in your calculations (.25% = .0025 in decimal form).

 

[ravenprp]

To calculate the amount of ore needed, we can use the following equation:

Mass of silver in ore = Mass of silver in 2 pounds of silver

We know that 2 pounds of silver is equivalent to 908 g (454 g x 2). Since the ore is 0.25% silver by mass, we can set up the equation as:

0.25% x Mass of ore = 908 g

To solve for the mass of ore, we need to convert 0.25% to decimal form by dividing it by 100:

0.0025 x Mass of ore = 908 g

Dividing both sides by 0.0025, we get:

Mass of ore = 908 g / 0.0025 = 363200 g

Now, we know that 1 cm^3 of the ore has a mass of 7 g. So, to convert the mass of ore from grams to cm^3, we can divide by the density of the ore:

Volume of ore = 363200 g / 7 g/cm^3 = 52000 cm^3

Therefore, 52000 cm^3 of ore with 0.25% silver by mass must be processed to obtain 2 pounds of silver. This is equivalent to 520 cm^3 or 0.52 L.

It is important to note that the density of the ore is given in g/cm^3, not cm^3/g. This may have caused the discrepancy in your calculation. Also, make sure to double check your units throughout the calculation to ensure accuracy.