Answer:
Concentration of stock solution in g/L = 5.0g/L
Concentration of stock solution in mol/L = 0.00117 mol/L or 0.00117 M
Explanation:
Concentration in mol/L = (Concentration in g/L)/(Molar mass)
Mass of (NH₄)₃Fe(ox)₃.3H₂O stock = 125mg = 0.125g
Volume of stock solution required = 25mL = 0.025 L
Concentration in g/L = 0.125/0.025 = 5.0 g/L
Molar Mass of (NH₄)₃Fe(ox)₃.3H₂O = (NH₄)₃Fe(C₂O₄)₃.3H₂O = 428.0632 g/mol
Concentration in mol/L = 5/428.0632 = 0.00117 mol/L = 0.00117 M
Hope this helps!!!
The molecular formula of the compound is - C₁₁H₂₀O₂
formula mass is the sum of the products of the relative atomic masses of the elements by the number of atoms of the corresponding element
r.a.m -relative atomic mass
the elements making up the compound with their relative atomic masses are;
C - 12 r.a.m - 11 C atoms
H - 1 r.a.m - 20 H atoms
O - 16 r.a.m - 2 O atoms
the formula mass- (11 x r.a.m of C ) + (20 x r.a.m of H ) + ( 2 x r.a.m of O)
formula mass = (11 x 12) + (20 x 1) + (2 x 16)
= 132 + 20 + 32
= 184
formula mass of compound is 184
Answer:
0.005404 M
Explanation:
Since you added an excess of sodium carbonate you warrantied that all the 
in the sample reacted with it. So we can say that the insoluble lead (II) carbonate 
contains all the ions in the original sample.
The moles of are:
One mol of is required to form one mol of . So, the stoichiometric relationship between them is 1:1.
Knowing this, 0.00054 is also the number of moles of in the original sample.
So, the concentration of in the original sample is:
The answer is that "<span>Ag+</span> ion combines with the chloride ion to form a precipitate".
Silver chloride is formed in the form of precipitates, Ag⁺ is a cation and Cl⁻ is an anion so both ions attract each other and combines to form AgCl.
The equation for the ionic reaction that occurs is;
<span><span>Ag</span></span>⁺(aq)+Cl⁻(aq)→AgCl(s)
