The number of Ml of a 0.40 %w/v solution of ,nalorphine that must be injected to obtain a dose of 1.5 mg is calculated as below
since M/v% is mass of solute in grams per 100 ml
convert Mg to g
1 g = 1000 mg what about 1.5 mg =? grams
= 1.5 /1000 = 0.0015 grams
volume is therefore = 100 ( mass/ M/v%)
= 100 x( 0.0015/ 0.4) = 0.375 ML
One would be phosporous whose configuration is 1s2 2s2 2p6 3s2 3p3
Answer: It would be Sp2, because H3O+ has planar structure.
Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Covalent
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For N₂,
E.N of Nitrogen = 3.04
E.N of Nitrogen = 3.04
________
E.N Difference
0.00 (Non Polar Covalent)
For Na₂O,
E.N of Oxygen = 3.44
E.N of Sodium = 0.93
________
E.N Difference 2.51 (Ionic)
For CO₂,
E.N of Oxygen = 3.44
E.N of Carbon = 2.55
________
E.N Difference 0.89 (Polar Covalent)
Answer:
B.
He could breed his peach plants with peach plants that can survive the frost.
Explanation:
Tom can best solve his problem by breeding his peach plants with the ones that can survive the frost.
- This is known cross breeding
- The goal of cross breeding is to produce offspring that contains the traits of both of the parents or an improved hybrid.
- If Tom can cross breed the failing peaches with the successful winter resistant ones, he can ensure better production.
- The new hybrid will therefore will be genetically superior and adapted to the prevailing environmental changes.
