Let us assume that there is a 100g sample of Opal. The masses of each element will be:
29.2g Si
33.3g O
37.5g H2O
Now we divide each constituent's mass by its Mr to get the moles present
Si: (29.2 / 28) = 1.04
O: (33.3 / 16) = 2.08
H2O: (37.5 / 18) = 2.08
Now we divide by the smallest number and obtain:
Si: 1
O: 2
H2O: 2
Thus, the empirical formula of Opal is:
SiO2 . 2H2O
The answer is C.
H₂O₂ ----> H₂O + O₂
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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Answer:
3.47 ×10^-10
Explanation:
The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)
A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.
E°cell = E°cathode - E°anode
E°cathode = -0.41 V
E°anode = -0.13 V
E°cell = -0.41 -(-0.13) = -0.28 V
From
E°cell = 0.0592/n log K
n= 2, K= the unknown
-0.28 = 0.0592/2 log K
log K = -0.28/0.0296
log K = -9.4595
K = Antilog ( -9.4595)
K= 3.47 ×10^-10
Answer:
C) sp2 and sp2
Explanation:
The hybridization depens on the ammount and type of bonds the atom analized has in the molecule.
For example:
- A C atom bonded to 4 H atoms has a sp3 hybridization.
- A C atom bonded to 2 H atoms and to 1 C with a double bond (like in ethene) has a sp2 hybridization
- A C bonded to 1 H and 1 C with a triple bond (like in ethyne) has a sp hybridization.
Analyzing the type and amount of unions of the nitrogen and the carbonyl you will be able to determine the hybridization.
In the imine, the N atom has a double bond to a C and a simple bond two other C, plus the lone pair of electrons (counts as a bond) so it will have a sp2 hybridization.
In the carbonyl, the C has two simple bonds to other C and a double bond to an oxygen atom. It will also have a sp2 hybridization
