Answer:
2.387 mol/L
Explanation:
The reaction that takes place is:
- 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O
First we <u>calculate how many moles of each reagent were added</u>:
- HCl ⇒ 200.0 mL * 3.85 M = 203.85 mmol HCl
- Ba(OH)₂ ⇒ 100.0 mL * 4.6 M = 460 mmol Ba(OH)₂
460 mmol of Ba(OH)₂ would react completely with (2*460) 920 mmol of HCl. There are not as many mmoles of HCl so Ba(OH)₂ will remain in excess.
Now we <u>calculate how many moles of Ba(OH)₂ reacted</u>, by c<em>onverting the total number of HCl moles to Ba(OH)₂ moles</em>:
- 203.85 mmol HCl *
= 101.925 mmol Ba(OH)₂
This means the remaining Ba(OH)₂ is:
- 460 mmol - 101.925 mmol = 358.075 mmoles Ba(OH)₂
There are two OH⁻ moles per Ba(OH)₂ mol:
- OH⁻ moles = 2 * 358.075 = 716.15 mmol OH⁻
Finally we <u>divide the number of OH⁻ moles by the </u><u><em>total</em></u><u> volume</u> (100 mL + 200 mL):
- 716.15 mmol OH⁻ / 300.0 mL = 2.387 M
So the answer is 2.387 mol/L
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.
To learn more about electrical energy visit:
brainly.com/question/863273?referrer=searchResults
B. Add 233 g of KCl to a 3.5 L container; then add enough water to dissolve the KCl and fill the container to the 3.5 L mark.
Answer:
Explanation:
Step 1. Determine the cell potential
<u> E°/V </u>
2×[Cr ⟶ Cr³⁺ + 3e⁻] 0.744 V
<u>3×[Cu²⁺ + 2e⁻ ⟶ Cu] </u> <u>0.3419 V
</u>
2Cr + 3Cu²⁺ ⟶ 3Cu + 2Cr³⁺ 1.086 V
Step 2. Calculate ΔG°
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