I believe that the answer is A i could be wrong though.
Answer:positive electrode
Explanation:
Anode can also be referred to as positive electrode in a cell
The reaction between NaOH and HCl is as follows
NaOH + HCl ---> NaCl + H₂O
for neutralisation, H⁺ ions react with an equivalent amount of OH⁻ ions.
Number of NaOH moles reacted = 0.270 M/1000 mL/L x 37 mL = 0.00999 mol
number of HCl moles reacted = 0.270 M/1000 mL x 27 mL = 0.00729 mol
HCl reacts with NaOH in 1:1 molar ratio
Number of excess NaOH moles remaining - 0.00999 - 0.00729 = 0.0027 mol
total volume of solution = 37 mL + 27 mL = 64 mL = 0.064 L
Since there's excess OH⁻ ions, we can calculate pOH value first
pOH = - log [OH⁻]
[OH⁻] = 0.0027 mol / 0.064 L = 0.042 mol/L
pOH = -log(0.042 M)
pOH = 1.37
by knowing pOH we can calculate pH using the following equation;
pH + pOH = 14
pH = 14 - 1.37
pH = 12.63
To determine the mass of the sample in milligrams in this problem, we use the avogadro's number to convert from atoms to moles, relate the moles of element in the sample to the mole present and the molar mass of the sample. We do as follows:
1.552 x 10^22 atoms H ( 1 mol H / 6.022x10^23 atoms H ) ( 1 mol C2H4Cl2 / 4 mol H ) ( 98.96 g C2H4Cl2 / 1 mol C2H4Cl2 ) = 0.625 g C2H4Cl2 = 625 mg <span>C2H4Cl2</span>
The answer is: 4) "slower, because the organic particles are molecules."
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