Sr is the limiting reactant.
Given the reaction equation;
2Sr + O2 (g) → 2SrO
2 moles of Sr reacts with 1 mole of O2
2 moles Sr will react with x mole of O2
x = 2 ×1/2
x = 1 mole of O2
Since we have more O2 than required, it is the reactant in excess, hence Sr is the limiting reactant.
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4. 2Li + 2H2O -> 2LiOH + H2
5. C6H12O6 + 6O2 -> 6CO2 + 6H2O
6. Zn + 2HCl -> ZnCl2 + H2
9. H2SO4 + Pb -> PbSO4 + H2
10. Ca(OH)2 + NH4Cl -> NH4 + CaCl2 + H2O
thats all i know
The result of the procedure is that the copper strip should go through oxidation.
<h3>How to explain the procedure?</h3>
The copper strip should be considered as the anode and the nail should be considered as the cathode.
The anode is the electrode of an electrochemical cell at which oxidation occurs. The anode should be + while the cathode should be negative. Both should be placed in the tank that should be filled with the electrolyte solution.
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The answer is the second one 2.3 X 10-3 dm3
Answer:
Molarity of 40 ml of NaoH is 0.3125 mL
Explanation:
As we know
Molarity of acid * volume of acid = molarity of base * volume of base
Substituting the given values, we get
Molarity of 40 ml of NaoH is 0.3125 mL
