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Answer: Na2CO3 + 2 HCl → 2 NaCl + H2CO3
I hope this helped!
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- Zack Slocum
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The concentration of [Sn⁺²] will be calculated by first calculating the moles of SnCl₂ added as these moles will give us the moles of [Sn⁺²] ion.
Moles of SnCl₂ = molarity X volume = 0.04 X 2.60 = 0.104 milli moles [as volume is in mL]
The moles of [Sn⁺² = 0.104 mmol
the total volume in solution = volume due to MO + volume due to SnCl₂ + volume due to HCl + volume due to NaCl
Total volume = 8+2.60+5.43+3.73= 19.76 mL
Concentration = moles / volume
concentration [Sn⁺²] = 0.104mmol / 19.76 mL = 0.0053 mol / L
Answer:positive electrode
Explanation:
Anode can also be referred to as positive electrode in a cell