Answer:
1.55 × 10²⁴ atoms Xe
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[Given] 57.5 L Xe at STP
[Solve] atoms Xe
<u>Step 2: Identify Conversions</u>
[STP] 22.4 L = 1 mol
Avogadro's Number
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Divide/Multiply [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
1.54583 × 10²⁴ atoms Xe ≈ 1.55 × 10²⁴ atoms Xe
Answer:
Write this in a word and skeleton equation:
Solid silver chloride and an aqueous solution of nitric acid are produced when a solution of silver nitrate is reacted with a solution of hydrochloric acid.
Explanation:
KE = 0
<h3>Further explanation </h3>
Energy is the ability to do work
Energy because its motion is expressed as Kinetic energy (KE) which can be formulated as:

So for two objects that have the same speed, the greater the mass of the object, the greater the kinetic energy
The stone in hand is in a motionless state (at rest) so that its velocity (v) = 0, so it has no kinetic energy
But this stone can have <em>potential energy that is gained due to its height</em>
Answer:
See explaination
Explanation:
1)
we know that
half cell with higher reduction potential is cathode
so
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
anode :
Cr(s) ---> Cr+3 + 3e-
so
overall reaction is
3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3
now
Eo cell = Eo cathode - Eo anode
so
EO cell = 1.77 + 0.74
Eo cell = 2.51 V
now
in this case
oxidizing agents are N20 and Cr+3
reducing agents are Cr and N2
higher the reduction potential , stronger the oxidizing agent
lower the reduction potential , stronger the reducing agent
so
oxidzing agents
N20 > Cr+3
reducing agents
Cr > N2
2)
cathode :
Au+ + e- --> Au
anode :
Cr ---> Cr+3 + 3e-
overall reaction
3Au+ + Cr ---> 3Au + Cr+3
Eo cell = 1.69 + 0.74
Eo cell = 2.43
now
oxidizing agents :
Au+ > Cr+3
reducing agents :
Cr > Au
3)
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
andoe :
Au ---> Au+ + e-
overall
2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20
Eo cell = 1.77 - 1.69
Eo cell = 0.08
oxidizing agents
N20 > Au+
reducing agents
Au > N2