Answer:
Please see the complete formt of the question below
Chlorine gas can be made from the reaction of manganese dioxide with hydrochloric acid.
MnO₂(s) + HCl(aq) → MnCl₂(aq) + H₂O(l) + Cl₂(g)
According to the above reaction, determine the limiting reactant when 5.6 moles of MnO₂ are reacted with 7.5 moles of HCl.
The answer to the above question is
The limiting reactant is the MnO₂
Explanation:
To solve this, we note that one mole of MnO₂ reacts with one mole of HCl to produce one mole of MnCl₂, one mole of H₂O and one mole of Cl₂
Molar mass of MnO₂ = 86.9368 g/mol
Molar mass of HCl = 36.46 g/mol
From the stoichiometry of the reaction, 5.6 moles of MnO₂ will react with 5.6 moles of HCl to produce 5.6 moles of H₂O and 5.6 moles of Cl₂
However there are 7.5 moles of HCL therefore there will be an extra 7.5-5.6 or 1.9 moles of HCl remaining when the reaction is completed
Those Hydrogen atoms which are present at alpha position to carbonyl group are mildly acidic in nature. When such acidic proton containing carbonyl compounds are treated with strong base, they yield enolates. The negative charge created on alpha carbon resonates and shifts to carbonyl oxygen resulting in formation of carbon double bond carbon.
In <span>tert−butyl methyl ketone there are two carbons at each alpha position. Among these two carbons only methyl carbon contains hydrogen atoms while the second one is bonded to further three carbons making it Quaternary carbon. The base abstracts proton from methyl group and enolate is formed.
</span>tert−butyl methyl ketone in this case acts as acid, Hydroxyl ion acts as base, while the enolate generated is the conjugate base of <span>tert−butyl methyl ketone and Water produced is the conjugate acid of hydroxide ion.</span>
Answer:
A basic oxide is an oxide which when combined with water gives off a base. When a substance reacts chemically, both as a base or acid it termed as an amphoteric solution. Neutral Oxide is one which neither has an acidic characteristic or a basic one. Metal Oxides have an oxidation number of -2 and generally comprise of an oxygen anion.
Answer:
b. Two moles of thiosulfate anion needed to react with one mole of hypochlorite anion.
Explanation:
1. Hypochlorite with iodide
2I⁻ ⟶ I₂ + 2e⁻
<u>ClO⁻ + 2H⁺ + 2e⁻ ⟶ Cl⁻ + H₂O </u>
2I⁻ + ClO⁻ + 2H⁺ ⟶ I₂ + Cl⁻ + H₂O
2. Thiosulfate with iodine
2S₂O₃²⁻ ⟶ S₄O₆²⁻ + 2e⁻
<u> I₂ + 2e⁻ ⟶ 2I⁻ </u>
2S₂O₃²⁻ + I₂ ⟶ S₄O₆²⁻ + 2I⁻
3. Sum of the reactions
2I⁻ + ClO⁻ + 2H⁺ ⟶ I₂ + Cl⁻ + H₂O
<u> 2S₂O₃²⁻ + I₂ ⟶ S₄O₆²⁻ + 2I⁻ </u>
2I⁻ + ClO⁻ + 2S₂O₃²⁻ + 2H⁺ ⟶ I₂ + Cl⁻ + S₄O₆²⁻ + H₂O
4. Molar ratio
S₂O₃²⁻:ClO⁻ = 2:1
Answer:
The artifact is 570 years old. That is, 5.7 × 10² years.
Explanation:
Radioactive decay follows first order reaction kinetics.
Let the initial activity for fresh Carbon-14 be A₀
And the activity at any other time be A
The rate of radioactive decay is given by
dA/dt = - KA
dA/A = - kdt
Integrating the left hand side from A₀ to A₀/2 and the right hand side from 0 to t(1/2) (where t(1/2) is the radioactive isotope's half life)
In [(A₀/2)/A₀] = - k t(1/2)
In (1/2) = - k t(1/2)
- In 2 = - k t(1/2)
k = (In 2)/t₍₁,₂₎
t(1/2) is given in the question to be 5.73 × 10³ years
k = (In 2)/5730 = 0.000121 /year
dA/A = - kdt
Integrating the left hand side from A₀ to A and the right hand side from 0 to t
In (A/A₀) = - kt
A/A₀ = e⁻ᵏᵗ
A = A₀ e⁻ᵏᵗ
A = 2.8 × 10³ Bq.
A₀ = 3.0 × 10³ Bq.
2.8 × 10³ = 3.0 × 10³ e⁻ᵏᵗ
0.9333 = e⁻ᵏᵗ
e⁻ᵏᵗ = 0.9333
-kt = In 0.9333
- kt = - 0.06899
t = 0.06899/0.000121 = 570.2 years = 5.7 × 10² years
