Answer: With the cold front, warm air is rapidly forced upward (like the shavings) in advance of the actual front (the “cutter”), creating towering cumulus clouds, some hard showers and quite possibly a few gusty thunderstorms followed by a push of cooler and drier air in its wake.
Explanation:
Hey there!:
Concentration of NaOH = 0.200 M
Concentration of HNO₃= 0.200 M
Total volume = 50.0 mL + 60.0 mL = 110 mL=> 0.11 L
The neutralization reaction between NaOH and HNO3 :
OH⁻ + H⁺ ----------> H₂O
So :
n ( H⁺ ) = 60 mL * 0.200 M / 1000 mL => 0.012 moles of H⁺
n ( OH⁻ ) = 50 mL 0.200 M / 1000 mL => 0.01 moles of OH⁻
Hence OH⁻ is limiting reagent .
Remaining moles of H⁺ = 0.012 - 0.01 => 0.002 moles
Concentration of H⁺ = 0.002 M / 0.11 L
Concentration of H⁺ = 0.01818 moles/L
Therefore:
pH = - log [ H⁺ ]
pH = - log [ 0.01818 ]
pH = 1.74
Hope that helps!
Answer:
Oxidation states are used in chemistry solutions. It is a bond in which electron transfers easily from one nucleus to another nucleus.
Explanation:
- Oxidation-reduction reactions have some rules.
- The oxidation state is 0 at an uncombined bond.
- The bond of oxidation reduction is +1 in alkeli metal.
- The bond in two metal is +2
- The oxidation reduction state at helogens is -1. It does not happened always.
- The oxygen bond in oxidation and reduction is -2.
- The sum of the oxidation state is equal to the compound charges.
- In this process the changes occur for any elements. Redox could be occur. Its oxidized and reduction reaction can be seen in this process.
It would be none would it not
Answer:
The correct answer is option A
