Hybridization refers to the mixing of atomic orbitals in an atom. The number of hybrid orbitals needs to be equal to the number of orbitals that have involved in prior to mixing.
The isolated atoms cannot prevail in a hybridized state as the atom in an isolated state do not form any kind of bond with the other atom, due to which the atomic orbitals do not go through the process of hybridization.
A. Quantity of saline = 500mL
Rate of infusion = 80 mL / h
Infusion time = Quantity / Rate = 500 mL / (80 mL/hr) = 6.25 hr
b. Child weight = 72.6 lb = 32.93 kg
Medrol to be given = 1.5 mg per kg
Quantity of Medrol = 20 mg/mL
Dosage available = 20 mg/mL / 1.5 mg/kg = 13.33 kg/mL
Dosage according to body weight = 32.93 kg / 13.33 kg/mL = 2.47 mL
This method of quantitative determination of percent purity is titrimetric reactions. These reactions most commonly involve neutralization reactions between an acid and a base. Then, we look at the neutralization reaction:
H₂C₂O₄ + 2 NaOH ⇒ Na₂C₂O₄ + 2 H₂O
So, we do the stoichiometric calculations. The important data we should know is the molar mass of oxalic acid which is equal to 90 g/mol.
(0.2283 mol/L NaOH * 0.3798 L * 1 mol H₂C₂O₄/ 2mol NaOH * 90 g/mol H₂C₂O₄) ÷ 0.7984 g *100%
= 488%
This is impossible. The purity can't be more than 100%. Looking at our calculations and the balance reaction, all steps were done correctly. So, I think there is some typographical error in the given. The mass of the sample should be 7.984 g. Then, the answer would be 48.87% purity.
Answer:
110ml
Explanation:
<em>Using the dilution equation, C1V1 = C2V2</em>
<em>Where C1 is the initial concentration of solution</em>
<em>C2 is final concentration of solution</em>
<em>V1 is intital volume of solution</em>
<em>V2 is final volume of solution.</em>
From the question , C1=6M, C2=0.5M, V1=10ml, V2=?



volume of water added = final volume -initial volume
= 120-10
=110ml
let me know when u find out plz because i would like to know as well its one of my chemistry qustions in an assiment. :)