A) Fe⁰ ----> Fe⁺³ +3e⁻ oxidation | *2
b) <u>Cu⁺² + 2e⁻ -----> Cu⁰ reduction |*3</u>
c) 2Fe⁰ +3Cu⁺² -----> 2Fe⁺³ + 2Cu⁰
Answer:
1.8g
Explanation:
Initial volume = 43.5ml
Final volume = 49.4ml
Mass = 10.88g
Density = ?
Volume = Final volume - initial volume
= 49.4 - 43.5
= 5.9ml
Density = Mass/volume
Density = 10.88/5.9
= 1.8g/ml
Answer:
<u><em></em></u>
- <u><em>pOH = 0.36</em></u>
Explanation:
Both <em>potassium hydroxide</em> and <em>lithium hydroxide </em>solutions are strong bases, so you assume 100% dissociation.
<u>1. Potassium hydroxide solution, KOH</u>
- Volume, V = 304 mL = 0.304 liter
- number of moles, n = M × V = 0.36M × 0.304 liter = 0.10944 mol
- 1 mole of KOH produces 1 mol of OH⁻ ion, thus the number of moles of OH⁻ is 0.10944
<u>2. LIthium hydroxide, LiOH</u>
- Volume, V = 341 mL = 0.341 liter
- number of moles, n = M × V = 0.341 liter × 0.51 M = 0.17391 mol
- 1mole of LiOH produces 1 mol of OH⁻ ion, thus the number of moles of OH⁻ is 0.17391
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<u>3. Resulting solution</u>
- Number of moles of OH⁻ ions = 0.10944 mol + 0.17391 mol = 0.28335 mol
- Volume of solution = 0.304 liter + 0.341 liter = 0.645 liter
- Molar concentration = 0.28335 mol / 0.645 liter = 0.4393 M
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<u>4. </u><em><u>pOH</u></em>
← answer
Answer:
your answer its c step by step that the first group