167 mL
P1V1 = P2V2
P1 = .8 atm
V1 = 250 mL
P2 = 1.2 atm
Solve for V2 —> V2 = P1V1/P2
V2 = (0.8 atm)(250 mL) / (1.2 atm) = 167 mL
Hey there!:
HCl + MnO2 → MnCl2 + H2O + Cl2
* in HCl the oxidation state of Cl is -1 .
* on the product side the oxidation state is 0 .
* therefore Cl gains electrons .
* in MnO2 the oxidation state of Mn is +4
* in MnCl2 the oxidation state of Mn is +2
Therefore Mn loses electrons
Answer A
Hope That helps!
The image of the bonds are missing, so i have attached it.
Answer:
A) - Sigma bond
-Sp³ and Sp³
- None
B) - Sigma and pi bond
- Sp² of C and p of O
- p of C and P of O
Explanation:
A) For compound 1;
- the molecular orbital type is sigma bond due to the end-to-end overlapping.
- Atomic orbitals in the sigma bond will be Sp³ and Sp³
- Atomic orbitals in the pi bond would be nil because there is no pi bond.
B) For compound 2;
- the molecular orbital type is sigma and pi bond
-Atomic orbitals in the sigma bond would be Sp² of C and p of O
- The Atomic orbitals in the pi bond will be; p of C and p of O
Answer:
Explanation:
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