Answer:-
2328.454 grams
Explanation:-
Volume V = 18.4 litres
Temperature T = 15 C + 273 = 288 K
Pressure P = 1.5 x 10^ 3 KPa
We know universal Gas constant R = 8.314 L KPa K-1 mol-1
Using the relation PV = nRT
Number of moles of oxygen gas n = PV / RT
Plugging in the values
n = (1.5 x 10^3 KPa ) x ( 18.4 litres ) / ( 8.314 L KPa K-1 mol-1 x 288 K)
n = 11.527 mol
Now the balanced chemical equation for this reaction is
2KNO3 --> 2KNO2 + O2
From the equation we can see that
1 mol of O2 is produced from 2 mol of KNO3.
∴ 11.527 mol of O2 is produced from 2 x 11.527 mol of KNO3.
= 23.054 mol of KNO3
Molar mass of KNO3 = 39 x 1 + 14 x 1 + 16 x 3 = 101 grams / mol
Mass of KNO3 = 23.054 mol x 101 gram / mol
= 2328.454 grams
Zinc would be considered the strongest reducing agent.
<h3>Reducing agent</h3>
A reducing agent is a chemical species that "donates" one electron to another chemical species in chemistry (called the oxidizing agent, oxidant, oxidizer, or electron acceptor). Earth metals, formic acid, oxalic acid, and sulfite compounds are a few examples of common reducing agents.
Reducers have excess electrons (i.e., they are already reduced) in their pre-reaction states, whereas oxidizers do not. Usually, a reducing agent is in one of the lowest oxidation states it can be in. The oxidation state of the oxidizer drops while the oxidizer's oxidation state, which measures the amount of electron loss, increases. The agent in a redox process whose oxidation state rises, which "loses/donates electrons," which "oxidizes," and which "reduces" is known as the reducer or reducing agent.
Learn more about reducing agent here:
brainly.com/question/2890416
#SPJ4
<h3 />
Answer: Fundamental, Key, vital, crucial
Explanation:
Answer: The answer is C. The reaction system absorbs 22 kJ of energy from the surroundings.
Explanation: I just took the quiz and it said I got it right.
Answer:
V = 2.32 Liters
Explanation:
PV = nRT => V = nRT/P
n = 25.8g/122g/mole = 0.21 mole
R = 0.08206 L·atm/mol·K
T = 25.44°C + 273 = 298.44K
P = 2.22 atm (given in problem)
V = (0.21mol)(0.08206 L·atm/mol·K)(298.44K)/(2.22atm) = 2.32 Liters at 25.44°C & 2.22atm
