The number of liters of 3.00 M lead (II) iodide : 0.277 L
<h3>Further explanation</h3>
Reaction(balanced)
Pb(NO₃)₂(aq) + 2KI(aq) → 2KNO₃(aq) + PbI₂(s)
moles of KI = 1.66
From the equation, mol ratio of KI : PbI₂ = 2 : 1, so mol PbI₂ :
Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution
Where
M = Molarity
n = Number of moles of solute
V = Volume of solution
So the number of liters(V) of 3.00 M lead (II) iodide-PbI₂ (n=0.83, M=3):
12 thousandth is the same as 0.012, so to put something in scientific notation you move the decimal point to the left or right until it is between the first two numbers. So:
Moving the decimal for 0.012 to between the first two numbers, we get: 1.2
For every move of the decimal place to get there, we divide/multiply by 10 each time. In this case, we multiply by 10 each time.
To get to 1.2, we had to move the decimal 2 times, so the answer is 1.2×10^-2
When you move the decimal to the right, the value of the power is negative, and when you move it to the left it is positive.
Are there any options for this question
<span>Reaction: CI2 + H2O ----> HCIO + HCI
Oxidations states:
The oxitation state of Cl2 = 0, because the oxidation state of an atom alone or a molucule with one kind of atom is always 0.
The
oxidation state of Cl in HClO is +1 because the oxidation state of H is
+ 1, the oxidation state of O is - 2, and the molecule is neutral, so
+1 + 1 - 2 = 0
The oxidation state of Cl in HCl is - 1, because the oxidation state of H is +1 and the molecule is neutral, so - 1 + 1 = 0.
Also,
you shall remember that when an atom increases its oxidation state is
is oxidized and when an atoms reduces its oxidations state it is
reduced.
With that you conclude that the right option is the last statement: </span>Cl
has an oxidation number of 0 in Cl2. It is then reduced to CI- with an
oxidation number of –1 in HCl and is oxidized to Cl+ with an oxidation
number +1 in HClO.
Answer:
2445 L
Explanation:
Given:
Pressure = 1.60 atm
Temperature = 298 K
Volume = ?
n = 160 mol
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 08206 L.atm/K.mol
Applying the equation as:
1.60 atm × V = 160 mol × 0.08206 L.atm/K.mol × 298 K
<u>⇒V = 2445.39 L</u>
Answer to four significant digits, Volume = 2445 L
