Answer:
the relation of two different forms of the same substance (such as two allotropic forms of tin) that have a definite transition point and can therefore change reversibly each into the other — compare monotropy.
Sodium Chloride is a compound.
<span>a. Use PV = nRT and solve for n = number of mols O2.
mols NO = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols O2 to mols NO2. Do the same for mols NO to mols NO2. It is likely that the two values will not be the same which means one is wrong; the correct value in LR (limiting reagent) problems is ALWAYS the smaller value and the reagent producing that value is the LR.
b.
Using the smaller value for mols NO2 from part a, substitute for n in PV = nRT, use the conditions listed in part b, and solve for V in liters. This will give you the theoretical yield (YY)in liters. The actual yield at these same conditions (AY) is 84.8 L.
</span>and % will be 60%.
Answer:
The number of moles of Sr in one mole of Sr(HCO₃)₂ = 1 mole
The number of moles of H in one mole of Sr(HCO₃)₂ = 2 moles
The number of moles of C in one mole of Sr(HCO₃)₂ = 2 moles
The number of moles of O in one mole of Sr(HCO₃)₂ = 6 moles
Explanation:
The given chemical formula of the compound is Sr(HCO₃)₂
The number of atoms of Sr in the compound = 1
The number of atoms of H in the compound = 2
The number of atoms of C in the compound = 2
The number of atoms of O in the compound = 6
The number of atoms of each element present in each formula unit of Sr(HCO₃)₂ is proportional to the number of moles of each atom in one mole of Sr(HCO₃)₂
Therefore;
The number of moles of Sr in one mole of Sr(HCO₃)₂ = 1 mole
The number of moles of H in one mole of Sr(HCO₃)₂ = 2 moles
The number of moles of C in one mole of Sr(HCO₃)₂ = 2 moles
The number of moles of O in one mole of Sr(HCO₃)₂ = 6 moles.
