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frosja888 [35]
4 years ago
12

A student constructs an electrochemical cell. A diagram of the operating cell and the unbalanced ionic equation representing the

reaction occurring in the cell are shown below. The blue color of the solution in the copper half-cell indicates the presence of Cu2+ ions. The student observes that the blue color becomes less intense as the cell operates.
Identify the type of electrochemical cell represented by the diagram.
Chemistry
1 answer:
Dominik [7]4 years ago
8 0

Answer:

A voltaic cell

Explanation:

A voltaic cell is a device which converts chemical energy to electrical energy. The chemical reactions that take place inside the cell causes electrons to flow from anode to cathode hence, electricity is produced. A simple voltaic cell is made by placing two different metals in contact with an electrolyte separated by a salt bridge. The cathode is the negative electrode while the anode is the positive electrode. It is also called a galvanic cell.

In a voltaic cell having a copper/copper solution half cell, reduction occurs at the cathode. Hence, at the cathode copper II ions accept two electrons and become reduced to ordinary metallic copper. This causes the blue colour of the solution to become discharged (fade) as the cell continues to function.

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______ reactions release energy.
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4 0
3 years ago
3.0 cm x 4.0 cm x 1.0 cm<br><br>[?]cm^3​
CaHeK987 [17]

Explanation:

<em>Hi</em><em> </em><em>there</em><em>!</em><em>!</em>

<em>you</em><em> </em><em>asked</em><em> </em><em>to</em><em> </em><em>multiply</em><em> </em><em>these</em><em> </em><em>all</em><em> </em><em>right</em><em>,</em>

<em>you</em><em> </em><em>can</em><em> </em><em>simply</em><em> </em><em>multiply</em><em> </em><em>it</em><em> </em><em>,</em>

<em>=</em><em>3</em><em>cm</em><em> </em><em>×</em><em> </em><em>4</em><em> </em><em>cm</em><em> </em><em>×</em><em> </em><em>1</em><em>cm</em>

<em>=</em><em> </em><em>1</em><em>2</em><em>cm</em><em>^</em><em>2</em><em>×</em><em>1</em><em>cm</em><em> </em><em> </em><em> </em><em> </em><em>(</em><em>4</em><em>×</em><em>3</em><em>=</em><em>1</em><em>2</em><em>)</em>

<em>=</em><em> </em><em>1</em><em>2</em><em>cm</em><em>^</em><em>3</em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>(</em><em>1</em><em>2</em><em>×</em><em>1</em><em>=</em><em>1</em><em>2</em><em>)</em>

<em>Therefore</em><em>, </em><em> </em><em>the</em><em>answer is</em><em> </em><em>1</em><em>2</em><em> </em><em>cm</em><em>^</em><em>3</em><em>.</em>

<em><u>Hope it helps</u></em><em><u>.</u></em><em><u>.</u></em>

4 0
3 years ago
How many moles of KBr are present in 500 ml of a 0.8 M KBr solution?
faltersainse [42]

Answer:

2) 0.4 mol

Explanation:

Step 1: Given data

  • Volume of the solution (V): 500 mL
  • Molar concentration of the solution (M): 0.8 M = 0.8 mol/L

Step 2: Convert "V" to L

We will use the conversion factor 1 L = 1000 mL.

500 mL × 1 L/1000 mL = 0.500 L

Step 3: Calculate the moles of KBr (solute)

The molarity is the quotient between the moles of solute (n) and the liters of solution.

M = n/V

n = M × V

n = 0.8 mol/L × 0.500 L = 0.4 mol

4 0
3 years ago
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