Answer:
Water (H2O) is polar due to the bent shape of the molecule. The shape means most of the negative charge from the oxygen is one one side of the molecule and the positive charge of the hydrogen atoms is on the other side of the molecule. This is an example of polar covalent chemical bonding.
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Answer:
It means the chemical entity is a radical
Explanation:
When we talk of unsaturation, we are referring to the number of pi-bonds in a chemical entity. The alkane, alkene and alkyne organic family are used to as common examples to explain the term unsaturation.
While alkynes have 3 bonds, it must be understood that they have 2 pi bonds only and as such their degree of saturation is two.
In the case of an alkene, there is only one single pi bond and as such the degree of unsaturation is 1.
Now in this case, we have a fractional 0.5 degree of unsaturation alongside the 3 to make a total of 3.5. So what’s the issue here?
The fractional part shows that the chemical entity we are dealing with here is a radical. While the integer 3 shows that there are 3 pi-bonds, the half pi bond remaining tells us that there is a missing electron on one of the atoms involved in the chemical bonding and as such, the 1/2 extra degree of unsaturation tends to tell us this.
Kindly recall that a radical is a chemical entity within which we have at the least an unpaired electron.
Answer:
C
Explanation:
The concept behind, is mole ratio of Al:FeO
Answer:
a. Ksp = 4s³
b. 5.53 × 10⁴ mol³/dm⁹
Explanation:
a. Obtain an expression for the solubility product of AB2(S),in terms of s.
AB₂ dissociates to give
AB₂ ⇄ A²⁺ + 2B⁻
Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as
AB₂ ⇄ A²⁺ + 2B⁻
1 : 1 : 2
Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s
So, we have
AB₂ ⇄ A²⁺ + 2B⁻
[s] [s] [2s]
So, the solubility product Ksp = [A²⁺][B⁻]²
= (s)(2s)²
= s(4s²)
= 4s³
b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³
Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³
Substituting the value of s into the equation, we have
Ksp = 4s³
= 4(2.4 × 10³ mol/dm³)³
= 4(13.824 × 10³ mol³/dm⁹)
= 55.296 × 10³ mol³/dm⁹
= 5.5296 × 10⁴ mol³/dm⁹
≅ 5.53 × 10⁴ mol³/dm⁹
Ksp = 5.53 × 10⁴ mol³/dm⁹
Answer: C(s) + O2(g) --> CO2(g)12g (C) .... 50.8g (O2)................. initial amounts0g(C) .........18.8g(O2) ................. amounts when reaction completeThat means that C was the limiting reactant, and the amount of CO2 is based on the amount of carbon that burned. Covert 12 grams of carbon to moles. The moles of CO2 will be the same, since they are in a 1:1 mole ratio. Then convert the moles of CO2 to grams.12g C x (1 mol C / 12.0 g C) x (1 mol CO2 / 1 mol C) x (44.0g CO2 / 1 mol CO2) =44 g of CO2
