25 drops of acid is required to neutralize the 50.0 ml of 0.010m of NaOH in the experiment.
The equation of the reaction is;
NaOH(aq) + HCl(aq) ---------> NaCl(aq) + H2O(l)
We can use the titration formula;
CAVA/CBVB = NA/NB
CA= concentration of acid
VA = volume of acid
CB = concentration of base
VB = volume of base
NA = number of moles of acid
NB = number of moles of base
CB = 0.010 M
VB = 50.0 ml
CA = 0.50 M
VA = ?
NA = 1
NB = 1
Substituting values;
CAVANB = CBVBNA
VA = 0.010 × 50.0 × 1/ 0.50 × 1
VA = 1 ml
Since the total volume of acid used is 1 ml and each drop contains 0.040 ml
The number of drops required is 1ml/0.040 ml = 25 drops
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Answer:
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espero que esto ayude
Oxidation occurs at the anode, so your answer is (2) loss of electrons
Answer: 118.5 grams
Explanation:
Molarity is defined as the number of moles of solute dissolved per liter of the solution.
where,
n= moles of solute
= volume of solution in ml = 500 ml
moles = 0.75
moles of solute =
0.75 =
mass of 
= 118.5 grams
Thus mass of needed to prepare 500 mL of this solution iis 118.5 grams
