The percentage of Chromium in Chromium Oxide is calculated as follow,
Step 1: Calculate Molar mass of Cr₂O₃,
Cr = 51.99 u
O = 16 u
So,
2(51.99) + 3(16) = 103.98 + 48 = 151.98 u
Step 2: Secondly divide molar mass of only chromium with total mass of Cr₂O₃ and multiply with 100.
i.e.
=

× 100
=
68.41 %
So, the %age composition of chromium in chromium oxide is
68.41 %.
46 is the answer. because if you add 26 and 20 that is the mass
1. Determine if the ionic substances can break apart into ions.
- e.g. CaCO3 isn't very soluble, do it can't dissolve and dissociate. If it can't pop apart, no ions.
2. Swap the partners for all the other ions that you can get from step 1. You can skip pairings with the same charge - a + can't get close to another + to react.
3. Use solubility, acid/base, and redox rules to see if anything will happen with the ions in solution.<span />
Answer: 0.050M urea, 0.10M glucose, 0.2M sucrose, pure water
Explanation:
Vapor pressure refers to the ease with which a liquid substance is transformed into vapour. High vapour density implies that the liquid is easily transformed into gas. Pure water is expected to have the lowest vapour density since it is held by strong intermolecular forces in the liquid state. Urea is an organic liquid held by weak Van der Waals forces hence its extremely high vapor pressure.