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1 year ago

Metallic elements can be recovered from ores that are oxides, carbonates, halides, or sulfides. Give an example of each type.

1 answer:

6 0

Metallic elements which can be recovered from ores that are oxides, carbonates, halides, or sulfides are iron, zinc, silver and lead respectively.

Ores which is deposited in earth's crust which contain minerals and metals. Metals can be obtained economically and sold commercially.

<h3>How metals obtained from sulphide or carbonate ore? </h3>

As we get to know that it is easy to obtain metals from their oxides. So, firstly ores which is found in the form of carbonates and sulphide are converted into their oxides by using the process of calcination and roasting.

The metals which is in the middle of the activity series are moderately reactive. These metals are found in the crust of the earth is mainly found as oxides, sulphides, or carbonates.

A metal which can occur in the form of sulphide ore is lead.

A metal which can occur in the form of oxide ore is iron.

A metal which can occur in the form of carbonate ore is zinc.

A metal which can occur in the form of halides ore is silver.

Thus, we concluded that the metallic elements which can be recovered from ores that are oxides, carbonates, halides, or sulfides are iron, zinc, silver and lead respectively.

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Calculate the molarity of 13.1 g of NaCl in 727 mL of solution

valentina_108 [34]

Answer:

The molarity of the solution is 0,31 M

Explanation:

We calculate the weight of 1 mol of NaCl from the atomic weights of each element of the periodic table. Then, we calculate the molarity, which is a concentration measure that indicates the moles of solute (in this case NaCl) in 1000ml of solution (1 liter)

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LuckyWell [14K]

From the equation;
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JulsSmile [24]

The number of moles present in 29.5 grams of argon is 0.74 mole.

The atomic mass of argon is given as;

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The number of moles present in 29.5 grams of argon is calculated as follows;

39.95 g ------------------------------- 1 mole

29.5 g ------------------------------ ?

$= \frac{29.5}{39.95} \\\\= 0.74 \ mole$