Given what we know, we can confirm that in a voltaic cell, the anode loses electrons and is oxidized, meanwhile, the cathode is reduced by gaining electrons.
<h3 /><h3>What is a voltaic cell?</h3>
- It is described as an electrochemical cell.
- These cells use chemical reactions to produce electrical energy.
- During this reaction, an anode loses electrons, thus oxidizing.
- Meanwhile, the cathode gains electrons and is reduced.
Therefore, given the nature of the voltaic cell, we can confirm that during its reaction, the anode is oxidized by losing electrons while the cathode becomes reduced by gaining them.
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Answer:
Explanation:
What we need to do here is to determine the ratios by using the Rydberg equation starting with the transition to n1 = 1, 2,3, etc and see which one fits the data. Remember the question states that they are series and the wavelengths will be for increasing energy levels.
1/λ = Rh x ( 1/n₁² - 1/n₂²)
Lyman series ( n₁=1 and n₂= 2,3 etc) for the first two lines, the ratios will be:
1/λ₁ /1/λ₂ =(1/1 -1/ 2²) / (1/1 -1/ 3²) ⇒ 0.84 ≠ 0.74 (the first ratio)
For Balmer series n₁ = 2 and n₂ = 3,4,5, etc
1/λ₁ /1/λ₂ =(1/4 -1/3²) / (1/4 -1/4²) ⇒ 0.741 = 0.741 (match!)
Lets use the third line to check our answer:
1/λ₁ /1/λ₂ =(1/4 -1/3²) / (1/4 -1/5²) = 0.66
Answer:
0.0400 g for the example given below.
Explanation:
pH value is not provided, so we'll solve this problem in a general case and then we will use an example to justify it.
- By definition,
. - NaOH is a strong base, as it's a hydroxide formed with a group 1A metal, so it dissociates fully in water by the equation:
. - From the equation above, using stoichiometry we can tell that the molarity of hydroxide is equal to the molarity of NaOH:
. - Concentration of hydroxide is then equal to the ratio of moles of NaOH and the volume of the given solution. Moles themselves are equal to mass over molar mass, so we obtain:
. - We also know that
. Take the antilog of both sides:
. - Solve for the mass of NaOH:
.
Now, let's say that pH is given as 12.00 and we use a 100-ml volumetric flask. Then we would obtain:
