Answer:
There were 0.00735 moles Pb^2+ in the solution
Explanation:
Step 1: Data given
Volume of the KI solution = 73.5 mL = 0.0735 L
Molarity of the KI solution = 0.200 M
Step 2: The balanced equation
2KI + Pb2+ → PbI2 + 2K+
Step 3: Calculate moles KI
moles = Molarity * volume
moles KI = 0.200M * 0.0735L = 0.0147 moles KI
Ste p 4: Calculate moles Pb^2+
For 2 moles KI we need 1 mol Pb^2+ to produce 1 mol PbI2 and 2 moles K+
For 0.0147 moles KI we need 0.0147 / 2 = 0.00735 moles Pb^2+
There were 0.00735 moles Pb^2+ in the solution
On temperature 25°C (298,15K) and pressure of 1 atm each gas has same amount of substance:
n(gas) = p·V ÷ R·T = 1 atm · 20L ÷ <span>0,082 L</span>·<span>atm/K</span>·<span>mol </span>· 298,15 K
n(gas) = 0,82 mol.
1) m(He) = 0,82 mol · 4 g/mol = 3,28 g.
d(He) = 10 g + 3,28 g ÷ 20 L = 0,664 g/L.
2) m(Ne) = 0,82 mol · 20,17 g/mol = 16,53 g.
d(Ne) = 26,53 g ÷ 20 L = 1,27 g/L.
3) m(CO) = 0,82 mol ·28 g/mol = 22,96 g.
d(CO) = 32,96 g ÷ 20L = 1,648 g/L.
4) m(NO) = 0,82 mol ·30 g/mol = 24,6 g.
d(NO) = 34,6 g ÷ 20 L = 1,73 g/L.
Answer:
A
Explanation:
Iron has the ground state electronic configuration [Ar]3d64s2
Fe2+ has the electronic configuration [Ar]3d6.
In an octahedral crystal field, there are two sets of degenerate orbitals; the lower lying three t2g orbitals, and the higher level two degenerate eg orbitals. Strong field ligands cause high octahedral crystal field splitting, there by separating the two sets of degenerate orbitals by a tremendous amount of energy. This energy is much greater than the pairing energy required to pair the six electrons in three degenerate orbitals. Since CN- is a strong field ligand, it leads to pairing of six electrons in three degenerate orbitals
When ice melts, the physicals state changes from solid to liquid. The energy or the heat required (q) required to change a unit mass (m) of a substance from solid to liquid is known as the enthalpy or heat of fusion (ΔHf). The variables; q, m and ΔHf are related as:
q = m * ΔHf
the mass of ice m = 65 g
the heat of fusion of water at 0C = ΔHf = 334 J/g
Therefore: q = 65 g * 334 J/g = 21710 J
Now:
4.184 J = 1 cal
which implies that: 21710 J = 1 cal * 21710 J/4.184 J = 5188.8 cal
Hence the heat required is 5188.8 cal or 5.2 Kcal (approx)
