It takes roughly 20,000 pounds of ore to produce one brick of gold weighing around 26 pounds. This high ratio accounts for the extensive processing and refining required to extract and purify the gold from the ore.
To calculate the amount of pure silver in the sample, multiply the mass of the ore by the percentage of silver: 0.53 grams * 5.4% = 0.02862 grams of silver. To convert grams to milligrams, multiply by 1000: 0.02862 grams * 1000 = 28.62 milligrams of pure silver in the sample.
The mass of iron ore is 6,085 kg.
From 1000 grams of ore containing 1 percent metal, you would extract 10 grams of metal. This is calculated by multiplying 1000 grams by 1 percent (0.01), which equals 10 grams.
The bulk density of iron ore pellets typically ranges from 2.2 to 2.6 grams per cubic centimeter.
To calculate this, you first need to determine the mass of pure iron in 454.0 grams. This is done by multiplying 454.0 grams by 0.35 (the decimal equivalent of 35.00%). This gives 158.9 grams of pure iron. To find out how many grams of ore is needed, you need to set up the equation 0.35 * Mass of ore = 158.9 grams, which gives you 454 grams of ore needed.
1 ton = 2000 pounds
The driving distance is about 580 miles.
Your Mother
It takes roughly 20,000 pounds of ore to produce one brick of gold weighing around 26 pounds. This high ratio accounts for the extensive processing and refining required to extract and purify the gold from the ore.
To calculate the amount of pure silver in the sample, multiply the mass of the ore by the percentage of silver: 0.53 grams * 5.4% = 0.02862 grams of silver. To convert grams to milligrams, multiply by 1000: 0.02862 grams * 1000 = 28.62 milligrams of pure silver in the sample.
70/0.39 = 179.49 grams = 0.18 kilograms (approx).
The mass of iron ore is 6,085 kg.
five pounds
1 ounce
The grade of the ore from which gold is to be recovered will determine how much of it must be processed to recover that ounce of gold. There is a broad variation, but there is plenty of gold in ores that is dispersed so widely that it is not economical to attempt commercial recovery. In an additional note, gold exists in veins within ores, and is not chemically combined with other elements within that ore. Mines that are probably considered marginal yield something on the order of 4 to 6 grams of gold per ton of ore, and you can do some simple math from there (perhaps using 5 grams/ton) to discover an answer for the stated grade of ore. There are just over 30 grams in a Troy ounce, and at 5 grams/ton, you're looking at 30 divided by 5 or 6 tons of ore to recover that (Troy) ounce of gold from ore with a grade of 5 grams/ton. If you have 6 tons of ore, that's a lot of ore, and you have a lot of hard work ahead of you if want the ounce of gold disbursed in it. It will take time and energy to win the metal, and it might be as cheap to buy the gold as to recover it from the ore. But we'll leave it for you to do the math after you get an assay.
yes/ but finding of gold is more presious than mining tons of ore