A chemist added an excess of sodium sulfate to a solution of a soluble barium compound to precipitate all of the barium ion as barium sulfate, BaSO4. How many grams of barium ion are in a 458-mg sample of the barium compound if a solution of the sample gave 513 mg BaSO4 precipitate

We have 302 mg barium ions in the solution, this is 65.9 %

Explanation:

Step 1: Data given

Mass of sample of the barium compound = 458 mg = 0.458 grams

A solution of the sample gave 513 mg BaSO4 precipitate

Molar mass of BaSO4 = 233.4 g/mol

Molar mass of Ba = 137.3 g/mol

Gravimetric factor  Ba/BaSO4=137.3/233.4=0.5883

Step 2: Calculate grams of barium ion

0.5883*513mg = 302mg or 0.302g

so there are 302mg Ba/458mg sol ution

%Ba=302mg Ba/458mg sol  *100= 65.9%Ba

We have 302 mg barium ions in the solution, this is 65.9 %

the answer is   c.) 1575 j.

explanation:

the term hydride is commonly named after binary compounds that hydrogen forms with other elements of the periodic table. hydride compounds in general form with almost any element, except a few noble gases. the trends and properties vary according to the type of intermolecular force that bonds the elements together, the temperature, its molecular masses, and other components. hydrides are classified into three major groups, depending on what elements the hydrogen bonds to. the three major groups are covalent, ionic, and metallic hydrides. formally, hydride is known as the negative ion of a hydrogen, h-, also called a hydride ion. because of this negative charge, hydrides have reducing, or basic properties. its special characteristics will be further discussed.

explanation: