<span>Answer:
Nothing is balanced in your final equation: not H, not O, not Cr, not I and your charges aren't either.
Start with your 2 half reactions:
I- --> IO3-
Cr2O72- --> 2 Cr3+
Balance O by adding H2O:
I- + 3 H2O --> IO3-
Cr2O72- --> 2 Cr3+ + 7H2O
Balance H by adding H+:
I- + 3 H2O --> IO3- + 6 H+
Cr2O72- + 14 H+ --> 2 Cr3+ + 7H2O
Balance charge by adding e-:
I- + 3 H2O --> IO3- + 6 H+ + 6 e-
Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7H2O
Since the numbers of electrons in your two half reactions are the same, just add them and simplify to give:
Cr2O72- + I- + 8 H+ --> IO3- + 2 Cr3+ + 4 H2O</span>
The Law of Conservation of Mass states that matter is not created nor destroyed.
Based on this, we can use some addition to find the answer.
3.00 + 1.40 = 4.40 g
Hope this helps :)
Answer:
V = 134.5 L
Explanation:
Given data:
Number of moles of KClO₃ = 4 mol
Litters of oxygen produced at STP = ?
Solution:
Chemical equation:
2KClO₃ → 2KCl + 3O₂
Now we will compare the moles of KClO₃ with oxygen.
KClO₃ : O₂
2 : 3
4 ; 3/2×4 = 6 mol
Litters of oxygen at STP:
PV = nRT
V = nRT/P
V = 6 mol × 0.0821 atm.L/mol.K × 273 K / 1atm
V = 134.5 L / 1
V = 134.5 L
Answer:
Substrate:alkyl halide
Leaving group: Cl
Organic product: The nitrile
Inorganic product: Cl-
Nucleophile: CN-
Explanation:
An SN2 reaction is a concerted bimolecular reaction. Concerted means that it involves two reactions taking place at the same time while bimolecular means that the rate determining step involves two molecules. The cyanide ion attacks the alkyl halide from the rear. In the transition state, the leaving group (Cl-) is departing while the nucleophile (CN-) is forming a bond to the alkyl halide simultaneously. The alkyl halide is the substrate in the reaction. The organic product is the nitrile shown in the image attached.
