Answer: 9.91×10²³ particles
Explanation:
To find the amount of particles, you will need to use the Ideal Gas Law with what we are given.
Ideal Gas Law: PV=nRT
After we find moles, we can use Avogadro's number to convert to particles.

P=101.3kPa=1.00 atm
V=4.0 L
T=23°C+273.15=296.15 K
R=0.08206 Latm/Kmol


Now that we have moles, we can convert to particles.

Answer:
It is just slightly less abundant than its alkali cousin, sodium. Potassium is less dense than water, so it can float on water. However, chemically, potassium reacts with water violently. It will give off hydrogen and eventually catch fire.
Answer:
In the given chemical reaction:
Species Oxidized: I⁻
Species Reduced: Fe³⁺
Oxidizing agent: Fe³⁺
Reducing agent: I⁻
As the reaction proceeds, electrons are transferred from I⁻ to Fe³⁺
Explanation:
Redox reaction is a chemical reaction involving the simultaneous movement of electrons thereby causing oxidation of one species and reduction of the other species.
The chemical species that <u><em>gets reduced by gaining electrons </em></u><u>is called an </u><u><em>oxidizing agent</em></u>. Whereas, the chemical species that <u><em>gets oxidized by losing electrons </em></u><u>is called a </u><u><em>reducing agent</em></u><u>.</u>
Given redox reaction: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂
<u>Oxidation half-reaction</u>: 2 I⁻ + → I₂ + 2 e⁻ ....(1)
<u>Reduction half-reaction</u>: [ Fe³⁺ + 1 e⁻ → Fe²⁺ ] × 2
⇒ 2 Fe³⁺ + 2 e⁻ → 2 Fe²⁺ ....(2)
In the given redox reaction, <u>Fe³⁺ (oxidation state +3) accepts electrons and gets reduced to Fe²⁺ (oxidation state +2) and I⁻ (oxidation state -1) loses electrons and gets oxidized to I₂ (oxidation state 0).</u>
<u>Therefore, Fe³⁺ is the oxidizing agent and I⁻ is the reducing agent and the electrons are transferred from I⁻ to Fe³⁺.</u>
P = 2.30 atm
Volume in liter = 2.70 mL / 1000 => 0.0027 L
Temperature in K = 30.0 + 273 => 303 K
R = 0.082 atm
molar mass O2 = 31.9988 g/mol
number of moles O2 :
P * V = n * R* T
2.30 * 0.0027 = n * 0.082 * 303
0.00621 = n * 24.846
n = 0.00621 / 24.846
n = 0.0002499 moles of O2
Mass of O2:
n = m / mm
0.0002499 = m / 31.9988
m = 0.0002499 * 31.9988
m = 0.008 g