Answer : The molecular geometry of is trigonal pyramidal.
Explanation :
Formula used :
where,
V = number of valence electrons present in central atom
N = number of monovalent atoms bonded to central atom
C = charge of cation
A = charge of anion
The given molecule is,
Number of bond pair = 3
Number of lone pair = 1
The total number of electron pair are 4 that means the hybridization will be and the electronic geometry of the molecule will be tetrahedral.
But as there are 3 atoms around the central oxygen atom, the fourth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be trigonal pyramidal.
Hence, the molecular geometry of is trigonal pyramidal.
CH3CH2CH2CH3 < CH3CH2CHO < CH3CHOHCH3
Explanation:
Boiling point trend of Butane, Propan-1-ol and Propanal.
Butane is a member of the CnH2n+2 homologous series is an alkane. Alkanes have C-H and C-C bonds which have Van der waals dispersion forces which are temporary dipole-dipole forces (forces caused by the electron movement in a corner of the atom). This bond is weak but increases as the carbon chain/molecule increases.
In Propan-1-ol(Primaryalcohol), there is a hydrogen bond present in the -OH group. Hydrogen bond is caused by the attraction of hydrogen to a highly electronegative element like Cl-, O- etc. This bond is stronger than dispersion forces because of the relative energy required to break the hydrogen bond. Alcohols (CnH2n+1OH) also experience van der waals dispersion forces on its C-C chain and C-H so as the Carbon chain increases the boiling point increases in the homologous series.
Propanal which is an Aldehyde (Alkanal) with the general formula CnH2n+1CHO. This molecule has a C-O, C-C and C-H bonds only. If you notice, the Oxygen is not bonded to the Hydrogen so there is no hydrogen bond but the C-O bond has a permanent dipole-dipole force caused by the electronegativity of oxygen which is bonded to carbon. It also has van der waals dispersion forces caused by the C-C and C-H as the carbon chain increases down the homologous series. The permanent dipole-dipole forces are not as easy to break as van der waals forces.
In conclusion, the hydrogen bonds present in alcohols are stronger than the permanent dipole-dipole bonds in the aldehyde and the van der waals forces in alkanes (irrespective of the carbon chain in Butane). So Butane < Propanal < Propan-1-ol
The answer for this question I think is D
Answer:
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Answer:
132g
Explanation:
The reaction equation is given as:
CaCO₃ → CaO + CO₂
Given;
Number of moles of CaCO₃ = 3 moles
Unknown:
Mass of CO₂ produced = ?
Solution:
From the balanced reaction equation;
1 mole of CaCO₃ will produce 1 mole of CO₂;
3 mole of CaCO₃ will then produce 3 moles of CO₂
To find the mass of CO₂;
Mass of CO₂ = number of moles x molar mass
Molar mass of CO₂ = 12 + 2(16) = 44g/mol
Mass of CO₂ = 3 x 44 = 132g