The mass of oxygen collected from the thermal decomposition of potassium chlorate at a temperature of 297 K and 762 mmHg is 0.16 g
<h3>How to determine the mole of oxygen produced </h3>
We'll begin by obtaining the number of mole of oxygen gas produced from the reaction. This can be obtained by using the ideal gas equation as illustrated below:
- Volume (V) = 0.128 L
- Temperature (T) = 297 K
- Pressure (P) = 762 – 22.4 = 739.6 mmHg
- Gas constant (R) = 62.363 mmHg.L/Kmol
- Number of mole (n) =?
PV = nRT
739.6 × 0.128 = n × 62.363 × 297
Divide both sides by 62.363 × 297
n = (739.6 × 0.128) / (62.363 × 297)
n = 0.0051 mole
Thus, the number of mole of oxygen gas produced is 0.0051 mole
<h3>How to determine the mass of oxygen collected</h3>
Haven obtain the number of mole of oxygen gas produced, we can determine the mass of the oxygen produced as follow:'
- Mole = 0.0051 mole
- Molar mass of oxygen gas = 32 g/mole
- Mass of oxygen =?
Mole = mass / molar mass
0.0051 = mass of oxygen / 32
Cross multiply
Mass of oxygen = 0.0051 × 32
Mass of oxygen = 0.16 g
Thus, we can conclude that the mass of oxygen gas collected is 0.16 g
Learn more about ideal gas equation:
brainly.com/question/4147359
#SPJ1
The electron is a subatomic particle that has a negative charge and a negligible mass. The electron travels around out side the nucleus.
The subatomic particles that are inside the nucleus are protons and neutrons.
The answer to your question is electron.
Answer:
Case 1 (energy level): In an atom, an electron jumps from energy level 1 to energy level 3. ... The energy will increase.
Explanation:
Explanation:
The given data is as follows.
Flow of chlorine (Q) = 
Amount of liter present per day is as follows.
It is given that dosage of chlorine will be as follows.
10 mg/l = 
kg/l
Therefore, total chlorine requirement is as follows.
Total chlorine requirement = 
kg/day
= 100 kg/day
Thus, we can conclude that the kilograms of chlorine used daily at the given water treatment plant is 100 kg/day.
Answer : The equilibrium concentration of 
will be, (C) 
Explanation : Given,
Equilibrium constant = 14.5
Concentration of 
at equilibrium = 0.15 M
Concentration of 
at equilibrium = 0.36 M
The balanced equilibrium reaction is,
The expression of equilibrium constant for the reaction will be:
![K_c=\frac{[CH_3OH]}{[CO][H_2]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCH_3OH%5D%7D%7B%5BCO%5D%5BH_2%5D%5E2%7D)
Now put all the values in this expression, we get:
![14.5=\frac{[CH_3OH]}{(0.15)\times (0.36)^2}](https://tex.z-dn.net/?f=14.5%3D%5Cfrac%7B%5BCH_3OH%5D%7D%7B%280.15%29%5Ctimes%20%280.36%29%5E2%7D)
![[CH_3OH]=2.82\times 10^{-1}M](https://tex.z-dn.net/?f=%5BCH_3OH%5D%3D2.82%5Ctimes%2010%5E%7B-1%7DM)
Therefore, the equilibrium concentration of will be, (C)
