Answer:
If we assume the molar volumes of water and ethanol 17.0 and 57.0 cm³/mol, respectively, Vmix = 20.5 cm³.
Explanation:
The molar volume of a substance is the ratio between the volume and the number of moles of the substance. It represents the volume that 1 mol of it occupies. Because we don't have access to page 24, let's assume the molar volumes of water and ethanol 17.0 and 57.0 cm³/mol, respectively.
The volume of mixture (Vmix) is the sum of the volume of each substance, which is the number of moles multiplied by molar volume, so:
Vmix = 0.300*57 + 0.200*17
Vmix = 17.1 + 3.4
Vmix = 20.5 cm³
Answer:
\large \boxed{\text{12.3 mol HCl}}
Explanation:
We need a balanced chemical equation with moles.
2HCl +Ca(OH)₂ ⟶ CaCl₂ + 2H₂O
n/mol: 12.3
The molar ratio is 2 mol H₂O:2 mol HCl.
<span>The answer is 4. The molecules of each material entice each other over dispersion (London) intermolecular forces. Whether a substance is a solid, liquid, or gas hinge on the stability between the kinetic energies of the molecules and their intermolecular magnetisms. In fluorine, the electrons are firmly apprehended to the nuclei. The electrons have slight accidental to stroll to one side of the molecule, so the London dispersion powers are comparatively weak. As we go from fluorine to iodine, the electrons are far from the nuclei so the electron exhausts can more effortlessly misrepresent. The London dispersion forces developed to be increasingly stronger.</span>
Answer:
½O 2 + 2e - + H 2O → 2OH.
Explanation:
Redox reactions - Higher
In terms of electrons:
oxidation is loss of electrons
reduction is gain of electrons
Rusting is a complex process. The example below show why both water and oxygen are needed for rusting to occur. They are interesting examples of oxidation, reduction and the use of half equations:
iron loses electrons and is oxidised to iron(II) ions: Fe → Fe2+ + 2e-
oxygen gains electrons in the presence of water and is reduced: ½O2 + 2e- + H2O → 2OH-
iron(II) ions lose electrons and are oxidised to iron(III) ions by oxygen: 2Fe2+ + ½O2 → 2Fe3+ + O2-
