Answer:
the answer is 5 m/s² and you can also divide jt by 2 to get its fully ans
Mole ratio :
<span>5 C</span>₆<span>H</span>₆<span>CHO + 2 KMnO</span>₄<span> + 6 H</span>⁺ <span>= 5 C</span>₆<span>H</span>₆<span>COOH + 2 Mn</span>²⁺<span> + 3 H</span>₂<span>O + 2 K</span>⁺
5 moles C₆H₆CHO ------------------ 2 moles KMnO₄
1.0 moles C₆H₆CHO ---------------- ( moles of KMnO₄ )
moles of KMnO₄ = 1.0 x 2 / 5
moles of KMnO₄ = 2 / 5
= 0.40 moles of KMnO4
hope this helps!
Br- + 2MnO4- + 2H+ → BrO3- + 2MnO2 + H2O
<h3>Further explanation</h3>
Given
MnO4- + Br- = MnO2 + BrO3-
Required
The half-reaction
Solution
In acidic conditions :
1. Add the coefficient
2. Equalization O atoms (add H₂O) on the O-deficient side.
3. Equalization H atoms (add H⁺ ) on the H -deficient side. .
4. Equalization of charge (add electrons (e) )
5. Equalizing the number of electrons and then adding the two half -reactions together
Oxidation : Br- → BrO3-
Reduction : MnO4- → MnO2
Br- + 3H2O → BrO3-
MnO4- → MnO2 + 2H2O
Br- + 3H2O → BrO3- + 6H+
MnO4- + 4H+ → MnO2 + 2H2O
Br- + 3H2O → BrO3- + 6H+ + 6e- x1
MnO4- + 4H+ + 3e- → MnO2 + 2H2O x2
- Equalizing the number of electrons and then adding the two half -reactions together
Br- + 3H2O → BrO3- + 6H+ + 6e-
2MnO4- + 8H+ + 6e- → 2MnO2 + 4H2O
Br- + 2MnO4- + 3H2O + 8H+ + 6e- → BrO3- + 2MnO2 + 6H+ + 4H2O + 6e-
Br- + 2MnO4- + 2H+ → BrO3- + 2MnO2 + H2O
The answer is d
The answer is d
Hello
We know both the elements involved in bonding are non-metals and the primary type of bonds involved in non-metals are covalent bonds. Covalent bonds are formed when two atoms share one or more electrons; thus we know that whatever the number of electrons shared, it has to be equal for both. We can eliminate choices A and B.
Next, we understand that it is easier for one atom to be shared among the two, rather than the 7. First, because Hydrogen needs only 1 electron to be stable and would require energy to compensate the remaining 6. Second, electrons are attracted towards the nucleus so it is inefficient to try and share 7 electrons when 1 provides the same amount of stability to Fluorine.
Therefore, the answer is C.