A group of students was asked to recover Cu(s) from a blue-g…
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A grоup оf students wаs аsked tо recover Cu(s) from а blue-green aqueous solution containing an unknown concentration of Cu2+(aq) . The students took a 100.0 mL sample of the solution and added an excess of 1.0 M Na3PO4(aq), causing the Cu2+(aq) to precipitate as Cu3(PO4)2(s), as shown in step 1 below. Step 1 The figure presents a diagram labeled Step 1. The diagram contains 3 beakers arranged horizontally. From left to right, the diagram indicates that the contents of the first two beakers are added together to produce the contents of the third beaker. The reaction is also represented by an equation written below the diagram. In the diagram, the first beaker is labeled blue green C u with a positive 2 charge, aqueous. This is added to the second beaker, which is labeled colorless 1.0 molar N a 3 P O 4, aqueous. The reaction of the contents of the first two beakers produces the contents of the third beaker, which is labeled blue green C u 3, open parenthesis, P O 4, close parenthesis, 2, solid. The equation written below the diagram is as follows: 3 C u with a positive 2 charge, aqueous, plus 2 N a 3 P O 4, aqueous, react to produce C u 3, open parenthesis, P O 4, close parenthesis, 2, solid, plus 6 N a with a positive one charge, aqueous. The Cu3(PO4)2(s) was filtered, dried, and weighed. Then the Cu3(PO4)2(s) was dissolved in a 3.0 M HCl(aq) solution, as shown in step 2 below. Step 2 The figure presents a diagram labeled Step 2. The diagram contains 3 beakers arranged horizontally. From left to right, the diagram indicates that the contents of the first two beakers are added together to produce the contents of the third beaker. The reaction is also represented by an equation written below the diagram. In the diagram, the first beaker is labeled blue green C u 3, open parenthesis, P O 4, close parenthesis, 2, solid. This is added to the second beaker, which is labeled colorless 3.0 molar H C l, aqueous. The reaction of the contents of the first two beakers produces the contents of the third beaker, which is labeled blue green C u C l 2, aqueous, plus H 3 P O 4, aqueous. The equation written below the diagram is as follows: C u 3, open parenthesis, P O 4, close parenthesis, 2, solid, plus 6 H C l, aqueous, react to produce 3 C u C l 2, aqueous, plus 2 H 3 P O 4, aqueous. The students added a strip of Zn(s) to the solution to recover the Cu(s), as shown in step 3 below. Step 3 The figure presents a diagram labeled Step 3. The diagram contains a beaker, a strip of metal, and a second beaker, arranged horizontally. The diagram indicates that the strip of metal is added to the first beaker to produce the contents of the second beaker. The reaction is also represented by an equation written below the diagram. In the diagram, the first beaker is labeled blue green C u C l 2, aqueous. The strip of metal is labeled silver colored Z n, solid. The reaction of the contents of the first beaker and the strip of metal produces the contents of the second beaker. The solution in the second beaker is labeled colorless Z n C l 2, aqueous, and material at the bottom of the beaker is labeled Brown C u, solid. The equation written below the diagram is as follows: C u C l 2, aqueous, plus Z n, solid, react to produce C u, solid, plus Z n C l 2, aqueous. Finally, the Cu(s) was filtered, dried, and weighed. If 3.8 g of Cu3(PO4)2(s) was recovered from step 1, what was the approximate [Cu2+] in the original solution? (The molar mass of Cu3(PO4)2 is 381 g/mol.)
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