Answer:
Pilot scale up
Explanation:
Preformulation studies are carried out on candidate drug molecules that show sufficient pharmacological promise in animal model(pharma approach).
It involves preliminary study of the properties of a drug which is considered a potentially active ingredient against a particular disease condition.
Scale-up is the term used to refer to the increase in the batch size of a product. This is only done after a drug has been proven successful against the target disease after extensive pilot studies.
Scale-up is the last operation carried out when a drug has passed through all stages. It is not included in preliminary preformulation studies
Answer:
Transition Element
Explanation:
Transition elements are defined as those elements which can form at least one stable ion and has partially filled d-orbitals. They are also characterized by forming complex compounds and having different oxidation states for a single metal element.
Transition metals are present between the metals and the non metals in the periodic table occupying groups from 3 to 12. There general electronic configuration is as follow,
(n-1)d
¹⁻¹⁰ns
¹⁻²
The general configuration shows that for a given metal, the d sublevel will be in lower energy level as compared to corresponding s sublevel. For example,
Scandium is present in fourth period hence, its s sublevel is present in 4rth energy level so its d sublevel will be present in 3rd energy level respectively.
Hence, we can conclude that for transition metals the electron are present in highest occupied s sublevel and a nearby d sublevel
.
Answer:
4 moles of H₃PO₄
Explanation:
The reaction expression is given as;
3KOH + H₃PO₄ → K₃PO₄ + 3H₂O
Number of moles of water = 12moles
Unknown:
Number of moles of H₃PO₄ = ?
Solution:
From the balanced reaction expression we see that;
3 moles of water is produced from 1 mole of H₃PO₄
So; 12 moles of water would be produced from = 4 moles of H₃PO₄
Find moles of MgSO4.7H2O
molar mass = 246
so moles = 32 / 246 = 0.13 moles.
When heated, all 7 H2O from 1 molecule will be gone.
total moles of H2O present = 7 x 0.13 = 0.91
mass of those H2O = 0.91 x 18 = 16.38g
so mass of anyhydrous MgSO4 remain = 32 - 16.38 = 15.62 g
I beleive its A but not very sure though. If im wrong sorry. But if im right I hope it helps (: