Answer:
Oxidation: Cr → Cr⁴⁺ + 4 e⁻
Reduction: 4 e⁻ + O₂ → 2 O²⁻
Explanation:
Let's consider the following redox reaction.
Cr + O₂ → CrO₂
Cr is oxidized. Its oxidation number increases from 0 to +4. The corresponding half-reaction is:
Cr → Cr⁴⁺ + 4 e⁻
O is reduced. Its oxidation number decreases from 0 to -2. The corresponding half-reaction is:
4 e⁻ + O₂ → 2 O²⁻
PH is your answer
Hope this helps!!
1. Translate, predict the products, and balance the equation above.
Li + Cu(NO3)2 = Li(NO3)2 + Cu
2. How many particles of lithium are needed to produce 125 g of copper?
125 g Cu ( 1 mol / 63.55 g ) (1 mol Li / 1 mol Cu ) ( 6.022 x 10^23 particles / 1 mol ) = 1.18x10^24 Li particles
3. How many grams of lithium nitrate are produced from 4.83E24 particles of copper (II) nitrate?
4.83E24 particles of copper (II) nitrate ( 1 mol / 6.022x10^23 particles ) (1 mol Li(NO3)2 / 1 mol Cu(NO3)2 ) ( 130.95 g / 1 mol ) = <span>1043.77 grams Li(NO3)2</span>
Answer:
The new temperature is 894 K or 621 °C
Explanation:
Step 1: Data given
Initial volume of the container = 2.000L
Initial temperature = 25.0 °C = 298 K
Volume increased to 6.00 L
Step 2: Calculate the new temperature
V1/T1 = V2/T2
⇒with V1 = the initial volume = 2.00L
⇒with T1 = the initial temperature = 25 °C = 298 K
⇒with V2 = the increased volume 6.00 L
⇒with T2 = the new temperature
2.00 L / 298 K = 6.00 L / T2
T2 = 894 K = 621 °C
The new temperature is 894 K or 621 °C
