The answer is (4) at the cathode, where reduction occurs. The Na+ gains one electron and become Na(l). So the reaction occurs at cathode and is reduction reaction.
Answer:
Evaporation
Explanation:
Though it doesn't require high temperatures to do so
False, they are all different because they help you know different things.
Answer:
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
Explanation:
Hello,
In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.
Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.
Regards.
Answer:
B.) 1.3 atm
Explanation:
To find the new pressure, you need to use Gay-Lussac's Law:
P₁ / T₁ = P₂ / T₂
In this equation, "P₁" and "T₁" represent the initial pressure and temperature. "P₂" and "T₂" represent the final pressure and temperature. After converting the temperatures from Celsius to Kelvin, you can plug the given values into the equation and simplify to find P₂.
P₁ = 1.2 atm P₂ = ? atm
T₁ = 20 °C + 273 = 293 K T₂ = 35 °C + 273 = 308 K
P₁ / T₁ = P₂ / T₂ <----- Gay-Lussac's Law
(1.2 atm) / (293 K) = P₂ / (308 K) <----- Insert values
0.0041 = P₂ / (308 K) <----- Simplify left side
1.3 = P₂ <----- Multiply both sides by 308
