2KOH + Cu(NO3)2 → 2KNO3 + Cu(OH)2
2K⁺ 1Cu²⁺ 2K⁺ 1Cu²⁺
2OH ⁻ 2NO3⁻ 2NO3⁻ 2OH⁻
Cu ions reduced to cu and Al oxidized to Al ions is the answer. copper ions is right hand side of the E.M.F cell where reduction take place and is reduced to copper metal. Al metal is at left hand side of the E.m.f cell where oxidation take place and is oxidized to Al ions
Answer:
potassium hydrogen phthalate KHP MOLAR MASS = 204.233 glmol
to get 1000 ml
Molar concentration = Mass concentration/Molar Mass
mass concentration = molar concentration x molar mass
mass concentration=0.1 M,
molar mass= 204.233 g/mol
so to get 1L
mass conc = 204.233 x 0.1
= 20.4233g for 1L or 1000 ml
to get 6.00 ml
if 20.4233g is for 1000ml
then to 6.00 ml
= 20.4233 x 6 / 1000
= 0.123g for 6.00 ml
according to the equation below
NaOH(aq) + KHC8H4O4(aq) --> KNaC8H4O4(aq) + H2O(l)
number of moles of NaOH is equal to that of KHP
so the same amount will be needed too, which is
= 0.123g
Answer:
0.12M
Explanation:
A balanced equation for the reaction will go a great deal in obtaining our desired result. So, let us write a balanced equation for the reaction
HCl + NaOH —> NaCl + H2O
From the above equation,
nA (mole of the acid) = 1
nB (mole of the base) = 1
Data obtained from the question include:
Vb (volume of the base) = 30mL
Mb (Molarity of the base) = 0.1M
Va (volume of the acid) = 25mL
Ma (Molarity of the acid) =?
The molarity of the acid can be obtained as follow:
MaVa/MbVb = nA/nB
Ma x 25/ 0.1 x 30 = 1
Cross multiply to express in linear form
Ma x 25 = 0.1 x 30
Divide both side by 25
Ma = (0.1 x 30) / 25
Ma = 0.12M
The molarity of the acid is 0.12M
