Answer:
H-O-H polar
O-C-O nonpolar
H-C-N polar
Explanation:
Looking up the electronegativities of the atoms involved in this question, we have:
Atom Electronegativity
H 2.2
C 2.55
N 3.04
O 3.44
All of the atoms differ in electronegativity resulting in individual dipole moments in H-O, O-C, H-C and C-N bonds. To find if the molecules will be polar we need to consider the structure of the compound to see if there is a resultant dipole moment.
In H-O-H, we have 2 lone pairs of electrons around the central oxygen atom which push the angle H-O-H of the ideal tetrahedral structure to be smaller than 109.5 º resulting in an overall dipole moment making it polar.
In O-C-O, we have two dipole moments that exactly cancel each other in the linear molecule since the central carbon atom does not have lone pairs of electrons since it has 2 double bonds. Therefore the molecule is nonpolar.
In H-C-N, again we have have a central carbon atom without lone pairs of electrons and the shape of the molecule is linear. But, now we have that the dipole moment in C-N is stronger than the H-C dipole because of the difference in electronegativity of nitrogen compared to hydrogen. The molecule has an overall dipole moment and it is polar.
Answer:
83.8%
Explanation:
The balanced reaction equation is;
2Al(s) + 3Cl2(g) → 2AlCl3(s)
Now we have to obtain the limiting reactant as the reactant that produces the least amount of AlCl3
Amount of Al = 3.11g/27 g/mol = 0.115 moles
If 2 moles of Al yields 2 moles of AlCl3
Then 0.115 moles of Al yields 0.115 moles of AlCl3
For Cl2
Amount of Cl2 = 5.32 g/71 g/mol= 0.075 moles
If 3 moles of Cl2 yields 2 moles of AlCl3
0.075 moles of Cl2 yields 0.075 * 2/3 = 0.05 moles of AlCl3
Hence Cl2 is the limiting reactant
Theoretical yield of AlCl3 = 0.05 moles of AlCl3 * 133g/mol = 6.65 g
%yield = actual yield /theoretical yield * 100
%yield = 5.57 g/6.65 g * 100
%yield = 83.8%
Answer:
ΔS° = 180.5 J/mol.K
Explanation:
Let's consider the following reaction.
4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)
The standard molar entropy of the reaction (ΔS°) can be calculated using the following expression.
ΔS° = ∑np × S°p - ∑nr × S°r
where,
ni are the moles of reactants and products
S°i are the standard molar entropies of reactants and products
ΔS° = 4 mol × S°(NO(g)) + 6 × S°(H₂O(g)) - 4 mol × S°(NH₃(g)) - 5 mol × S°(O₂(g))
ΔS° = 4 mol × 210.8 J/K.mol + 6 × 188.8 j/K.mol - 4 mol × 192.5 J/K.mol - 5 mol × 205.1 J/K.mol
ΔS° = 180.5 J/K
This is the change in the entropy per mole of reaction.
Answer:
a. Synthesis reaction
b. Single - Displacement
c. Decomposition reaction
d. Combustion Reaction
Explanation:
Synthesis reaction: Reaction in which two or more substance combine to give a single compound.
Single - Displacement : Reaction in which the more reactive element displaces the less reactive element.Here Li replace OH from water.
Decomposition reaction: Reaction in which a single reactant decompose to give two or more products.
Combustion Reaction : IT is a redox reaction in which a substance is burned in the presence of oxygen to give carbon dioxide and water.
A. Both the cell membrane and the nuclear membrane are protective coverings.