Balance the following redox equation and identify the element oxidized, the element reduced, the oxidizing agent, and the reduci
The charge of Br changed from –1 to 0, therefore it is the element which is oxidized. Since it is oxidized then Br is also the reducing agent.
The charge of Mn changed from +4 to +2 therefore it is the element which is reduced. Since Mn is reduced, then MnO2 is the oxidizing agent.
The half –reactions are:
Br: 2Br --> Br2 + 2e-
Mn: MnO2 --> Mn2+
First balance oxygen by adding H2O:
MnO2 --> Mn2+ + 2H2O
Then balance hydrogen by adding H+ ions:
4H+ + MnO2 --> Mn2 + 2H2O
Then the appropriate electrons:
4e- + 4H+ + MnO2 --> Mn2 + 2H2O
Multiply the half-reaction of Br by 2 because the half-reaction of Mn has 4 electrons.
4Br --> 2Br2 + 4e-
Combine the two half reactions and cancel common factors:
4Br- + 4H+ + MnO2 --> 2Br2 + Mn2 + 2H2O
When an electron quickly occupies an strength state increased than its ground state, it is in an excited state. An electron can end up excited if it is given greater energy, such as if it absorbs a photon, or packet of light, or collides with a close by atom or particle.

Explanation:
Mass = volume × density
Mass = 652 cm³ × 21.45 g/cm³
= 13985.4 g
Explanation:
Answer:
allows students to contribute to life-saving remedies, enhance the speed of delivery of new medications, and help others.
Answer:
The answer to your question is all the formulas in bold has the same empirical formula
Explanation:
Data
Empirical formula CH₂O
Process
To solve this problem factor the subscripts of each formula and compare the result with the empirical formula given.
a) C₂H₄O₂ factor 2 2(CH₂O)
b) C₃H₆O₃ factor 3 3(CH₂O)
c) CH₂O₂ this formula can not be simplified
d) C₅H₁₀O₅ factor 5 5(CH₂O)
e) C₆H₁₂O₆ factor 6 6(CH₂O)