Answer: Option (b) is the correct answer.
Explanation:
A covalent bond is defined as the bond which occurs due to sharing of electrons between the combining atoms.
Generally, a covalent bond is formed between non-metals.
For example, both nitrogen and oxygen atoms are non-metals and they combine covalently to form 
compound.
As nitrogen has 5 valence electrons and an oxygen atom has 6 valence electrons. So, there occurs unequal sharing of electrons between the two.
Thus, we can conclude that when a covalent bond forms then electrons in valence shells are shared between atoms.
Answer:
0.20 mol
Explanation:
Let's consider the reduction of iron from an aqueous solution of iron (II).
Fe²⁺ + 2 e⁻ ⇒ Fe
The molar mass of Fe is 55.85 g/mol. The moles corresponding to 5.6 g of Fe are:
5.6 g × 1 mol/55.85 g = 0.10 mol
2 moles of electrons are required to deposit 1 mole of Fe. The moles of electrons required to deposit 0.10 moles of Fe are
0.10 mol Fe × 2 mol e⁻/1 mol Fe = 0.20 mol e⁻
Answer:
Explanation:
From the given information:
The equation for the reaction can be represented as:
The I.C.E table can be represented as:
2SO₂ O₂ 2SO₃
Initial: 14 2.6 0
Change: -2x -x +2x
Equilibrium: 14 - 2x 2.6 - x 2x
However, Since the amount of sulfur trioxide gas to be 1.6 mol.
SO₃ = 2x,
then x = 1.6/2
x = 0.8 mol
For 2SO₂; we have 14 - 2x
= 14 - 2(0.8)
= 14 - 1.6
= 12.4 mol
For O₂; we have 2.6 - x
= 2.6 - 1.6
= 1.0 mol
Thus;
[SO₂] = moles / volume = ( 12.4/50) = 0.248 M ,
[O₂] = 1/50 = 0.02 M ,
[SO₃] = 1.6/50 = 0.032 M
Kc = [SO₃]² / [SO₂]² [O₂]
= ( 0.032²) / ( 0.248² x 0.02)
= 0.8325
Recall that; the equilibrium constant for the reaction = 0.8325;
If we want to find:
Then:
Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.
The right answer is= <span>rolling, fluid, and sliding</span>
You might have meant hemiacetal, not hemicetal.
Acetals contain two –OR groups, one –R group and a –H atom. In hemiacetals, one of the –OR groups in acetals is replaced by a –OH group<span>.
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