Answer:
The mole fraction is the number of moles of something divided by the number of total moles present. First, let us work out the number of moles of KCl present. A 26.3% w/w solution would contain 26.3 g of KCl per 100 g of solution. Hence, 26.3 g of KCl is 26.374.5513 = 0.353 moles
Explanation:
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The solubility constant or Ksp is calculated by the product of the concentration of the ions raise to the stoichiometric coefficient. We calculate as follows:
BaCrO4 = Ba2+ + CrO42-
Ksp = [Ba2+][CrO4]
Ksp = [1.08×10−5<span> ] [1.08×10−5] = 1.1664x10^-10</span>
I think the correct answer would be B. Octane is a component of fuel used in internal combustion engines. the dominant intermolecular forces in octane are london dispersion forces. Dipole-dipole bonds happens only with polar substances. Hydrogen bonding occurs when H bonds with an O, F or N atom. Covalent bonds are not intermolecular force rather it is an intramolecular force. Carbon-Hydrogen bonds, as far as I know, are not a type of bond, intermolecular or intramolecular. So, we are left with london dispersion force. It is a temporary force that happens when electrons of two atoms in adjacent would occupy positions that would form dipoles temporarily.
The limiting reactant is Cl₂ (chlorine).
Explanation:
We have the following chemical reaction:
2 Al (s) + 3 Cl₂ (g) → 2 AlCl₃ (s)
And we start with 2.7 g of Al and 3.12 g of Cl₂.
First we calculate the number of moles of each reactant.
number of moles = mass / molar weight
number of moles of Al = 2.7 / 27 = 0.1 moles
number of moles of Cl₂ = 3.12 / 71 = 0.044 moles
From the chemical reaction we see that 2 moles of Al are reacting with 3 moles of Cl₂ so 0.1 moles of Al are reacting with 0.15 moles of Cl₂ which is a quantity higher than our available one of 0.044 moles of Cl₂. The limiting reactant is Cl₂.
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Its B- a mixture.
Because I don't think you could separate and atom by distillation or filtration.
Actually I don't think you could separate an element by them.
I'm pretty sure compound can't be separated by them either.