The answer is (4) at the cathode, where reduction occurs. The Na+ gains one electron and become Na(l). So the reaction occurs at cathode and is reduction reaction.
Its obviously D like what are you stupid lol jk don’t take it to heart kid
The empirical formula of the compound obtained from the question given is NaBrO₃
<h3>Data obtained from the question </h3>
- Sodium (Na) = 15.24%
- Bromine (Br) = 52.95%
- Oxygen (O) = 31.81%
<h3>How to determine the empirical formula </h3>
The empirical formula of the compound can be obtained as illustrated below:
Divide by their molar mass
Na = 15.24 / 22.99 = 0.663
Br = 52.95 / 79.90 = 0.663
O = 31.81 / 16 = 1.988
Divide by the smallest
Na = 0.663 / 0.663 = 1
Br = 0.663 / 0.663 = 1
O = 1.988 / 0.663 = 3
Thus, the empirical formula of the compound is NaBrO₃
Learn more about empirical formula:
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Convert Mg to grams
1g =1000mg what about 3.91 Mg
= 3.91mg x 1g/1000mg= 3.91 x10^-3 g
moles= mass/molar mass
that is 3.91 x10^-3g /99 g/mol=3.95 x10^-5moles
concentration= moles / vol in liters
that is 3.95 x10^-5/100 x1000= 3.94 x10^-4M
equation for dissociation of CUCl= CUCl----> CU^+ +Cl^-
Ksp=(CU+)(CI-)
that is (3.95 x10^-4)(3.95 x10^-4)
Ksp= 1.56 x10^-7
Water molecules move througout the solute
