Answer:
Equation of reaction:
a) 2HCl + Ba(OH)2 ==> CaCl2 + 2H2O
b) Molarity of base = 0.042 M.
Explanation:
Using titration equation
CAVA/CBVB = NA/NB
Where NA is the number of mole of acid = 2
NB is the number of mole of base = 1
CA is the molarity of acid =0.15M
CB is the molarity of base = to be calculated
VA is the volume of acid = 25 ml
VB is the volume of base = 44.45mL
Substituting
0.15×25/CB×44.45 = 2/1
Therefore CB =0.15×25×1/44.45×2
CB = 0.042 M.
Answer:
<em>For both cases the answer is C</em>
Explanation:
We can see that the orbitals are not filled in the order of increasing energy and the Pauli exclusion principle is violated because it does not follow the correct order of the electron configuration; In the first exercise after the 2s2 orbital, the 2p2 orbital follows.
For the second exercise, you must start in order with level 1 and correctly filling each of the sublevels corresponding to each level until reaching level 7 and thus completing the desired number of electrons.
Mole=number of molecules/6.02x10²³
mole=2
number of molecules= 2x6.02x10²³
number of molecules=12.04x10²³
Empirical formula of ionic compound is FeO. In which the composition of atoms is 1 : 1.
Empirical formula of an ionic compound is defined as the formula which gives whole number ratio of atoms of various elements present in molecule of compund.
mass of iron in compound = 34.95 g
molar mass of iron = 55.8 g
mass of oxygen in compound = 15.05 g
molar mass of oxygen = 32 g
number of moles of iron present in the compound are ratio of mass of iron in compound/ molar mass of iron
number of moles of iron in compound= 34.95 / 55.8 = 0.6263 ~ 1
number of moles oxygen in compound= 15.05/ 32 = 0.473 ~ 0.5
the ratio of the number of oxygen atoms to number of iron atoms present in one formula unit of iron compund is 2×0.5 / 1 = 1 : 1
Hence , the required empirical formula of iron compound is FeO.
To learn more about Emiprical formula, refer:
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It would be MnSO4
The (II) lets you know it’s the form with a 2+ charge and Sulfate has a 2- charge
These will cancel out making it plain MnSO4
If it was manganese (iii) sulfide the answer would be Mn2(SO4)3
