Number of moles = 0.0688 moles of NaoH
volume = 0.250 L
Molarity = moles of solute / volume ( L )
M = 0.0688 / 0.250
M = 0.28 M
Answer B
The combustion of an organic compound is mostly written as,
CaHbOc + O2 --> CO2 + H2O
where a, b, and c are supposed to be the subscripts of the elements C, H, and O in the compound. Determining the number of moles of C and H in the product which is the same as that in the compound,
(Carbon, C) : (561 mg) x (12/44) = 153 mg x (1 mmole/12 mg) = 12.75
(Hydrogen, H) : (306 mg) x (2/18) = 34 mg x (1 mmole/1 mg) = 34
Calculating for amount of O in the sample,
(oxygen, O) = 255 - 153 mg - 34 mg = 68 mg x (1mmole/16 mg) = 4.25
The empirical formula is therefore,
C(51/4)H34O17/4
C3H8O1
The molar mass of the empirical formula is 60. Therefore, the molecular formula of the compound is,
C9H24O3
Given:
Stock dose/concentration of 20% Acetylcysteine (200 mg/mL)
150 mg/kg dose of Acetylcysteine
Weight of the dog is 13.2 lb
First we must convert 13.2 lb to kg:
13.2 lb/(2.2kg/lb) = 6 kg
Then we must calculate the dose:
(150 mg/kg)(6kg) = 900 mg
Lastly, we must calculate the dose in liquid form to be administered:
(900 mg)/(200 mg/mL) = 4.5 mL
Therefore, 4.5 mL of 20% Acetylcysteine should be given.
Answer:
D. The presence of a base catalyst.
E. The presence of an acid catalyst. and
F. The presence of heat.
Explanation:
