Answer:
Colors of transition metal compounds are due to two types of electronic transitions. Due to the presence of unpaired d electrons, transition metals can form paramagnetic compounds. Transition metals are conductors of electricity, possess high density and high melting and boiling points.
Explanation:
Answer:
a) yes, it was an hydrate
b) the number of waters of hydration, x = 6
Explanation:
a) yes it was an hydrate because the mass decreased after the process of dehydration which means removal of water thus some water molecules were present in the sample.
b) NiCl2. xH2O
mass if dehydrated NiCl2 = 2.3921 grams
mass of water in the hydrated sample = mass of hydrated - mass of dehydrated = 4.3872 - 2.3921 = 1.9951 g which represent the mass of water that was present in the hydrated sample.
NiCl2.xH2O
mole of dehydrated NiCl2 = m/Mm = 2.3921/129.5994 = 0.01846 mole
mole of water = m/Mm = 1.9951/18.02 = 0.11072 mole
Divide both by the smallest number of mole (which is for NiCl2) to find the coefficient of each
for NiCl2 = 0.01846/0.01846 = 1
for H2O = 0.11072/0.01846 = 5.9976 = 6
thus the hydrated sample was NiCl2. 6H2O
True. For example, electron domain geometry and molecular geometry of water and ammonia are different.
Answer:
The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol
Explanation:
Step 1: Data given
The combustion reaction of octane produces 5104.1 kJ per mol octane
Step 2: The balanced equation
C8H18(g) + 12.5 O2 ⟶ 8CO2 (g) + 9 H2O (g) ∆H°rxn = -5104.1 kJ/mol
Step 3:
∆H°rxn = ∆H°f of products minus the ∆H° of reactants
∆H°rxn = ∆H°f products - [∆H°f reactants]
-5104.1 kJ/mol = (8*∆H°fCO2 + 9*∆H°fH20) - (∆H°fC8H18 + 12.5∆H°fO2)
∆H°f C8H18 = ∆H°f 8CO2 + ∆H°f 9H2O+ 5104.1 kJ/mol
∆H°f C8H18 = 8 * (-393.5 kJ)/mol + 9 * (-241.8 kJ/mol)] + 5104.1 kJ
/mol
∆H°f C8H18 = -220.1 kJ/mol
The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol
