Answer:
e. 3.08 x 10⁻² mol of ions.
Explanation:
- Every 1.0 mole of any compound contains Avogadro's number of molecules (6.022 x 10²³).
- We can get the no. of moles of NiCl₂ using cross multiplication:
1.0 mol NiCl₂ contains → 6.022 x 10²³ molecules.
??? mol NiCl₂ contains → 6.188 x 10²¹ molecules.
∴ The no. of moles of NiCl₂ = (1.0 mol)(6.188 x 10²¹ molecules)/(6.022 x 10²³ molecules) = 1.028 x 10⁻² mol.
- NiCl₂ is ionized according to the equation:
NiCl₂ → Ni²⁺ + 2Cl⁻.
Which means that every 1.0 mol of NiCl₂ is ionized to produce 3.0 moles (1.0 mol of Ni²⁺ and 2 moles of Cl⁻).
<em>∴ The total moles of ions are released</em> = 3 x 1.028 x 10⁻² mol = <em>3.083 x 10⁻² mol of ions.</em>
Answer:
In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.
Explanation:
tally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:
\displaystyle \%\text{H}=\frac{\text{mass H}}{\text{mass compound}}\times 100\%%H=
mass compound
mass H
×100%
\displaystyle \%\text{C}=\frac{\text{mass C}}{\text{mass compound}}\times 100\%%C=
mass compound
mass C
×100%
If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:
\displaystyle \%\text{H}=\frac{2.5\text{g H}}{10.0\text{g compound}}\times 100\%=25\%%H=
10.0g compound
2.5g H
×100%=25%
\displaystyle \%\text{C}=\frac{7.5\text{g C}}{10.0\text{g compound}}\times 100\%=75\%%C=
10.0g compound
7.5g C
×100%=75%
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):
First you have a knowledge of bond order which is
B.O=(no. of electrons in bonding orbital - no. of electrons in non-bonding orbital)÷2
Note:
bond strength is directly proportional to bond order.
For oxygen:
B.O=(6-2)/2= 2; after the removal of two electrons(removal occur from non-bonding orbital)
B.O=(6-0)/2= 3 (As B.O increased bond strength increased)
For Nitrogen:
B.O=(6-0)/2= 3; after the removal of two electrons(removal occur from bonding orbital)
B.O=(4-0)/2= 2 (As B.O decreased bond strength decreased)
