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stepladder [879]
3 years ago
15

A certain metal forms a soluble nitrate salt M(NO3)3. Suppose the left half cell of a galvanic cell apparatus is filled with a 3

.0mM solution of M(NO3)3 and the right half cell with a 3.0M solution of the same substance. Electrodes made of M are dipped into both solutions and a voltmeter is connected between them. The temperature of the apparatus is held constant at 20.0 C.
Required:
a. Which electrode will be positive?
b. What voltage will the voltmeter show? Assume its positive lead is connected to the positive electrode.

Chemistry
1 answer:
DiKsa [7]3 years ago
7 0

Answer:

1.The electrode on the right is positive

2. 0.058V

Explanation:

The above cell is a concentration cell.

A concentration cell is an electrolytic cell that is made of two half-cells with the same electrodes, but differs in concentrations of the solutions. A concentration cell functions by diluting the more concentrated solution and concentrating the more dilute solution, creating a voltage as the cell reaches an equilibrium thereby transferring the electrons from the cell with the lower concentration to the cell with the higher concentration.

In the above cell, electrons flow from the left electrode (less concentrated) to the right electrode (more concentrated). Therefore, the right electrode is the positive electrode (cathode).

Part 2: Please, see the attachment below for the calculations.

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Explanation:

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How many grams of Cl are in .46 moles of Cl?
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given the following equation: 8 H2 + S8 −−> 8H2S what is the molar relationship between H2 and H2S
PSYCHO15rus [73]
-(1/8)Δ[H2S]/Δt = -(1/4)Δ[O2]/Δt
(1/8)Δ[H2S]/Δt = (1/4)Δ[O2]/Δt
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4 0
3 years ago
you ned to make an aqeous solution of .139M aluminum nitrate for an experiment in a lab, using 300. ml volumetric flask. how muc
larisa [96]

Answer:

8.88 grams of aluminum nitrate should be weighted.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to use the definition of molarity to calculate the moles of aluminum nitrate as follows:

M=\frac{n}{V}\\\\n=M*V\\\\n=0.139mol/L*0.300L=0.0417mol

Now, since the molar mass of aluminum nitrate is 212.996 g/mol, we obtain the following mass:

m=0.0417mol*\frac{212.996 g}{1mol} \\\\m=8.88g

Therefore, 8.88 grams of aluminum nitrate should be weighted.

Regards!

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What are the answer? For me to help you I need to see them
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