Answer:
P_2 =0.51 atm
Explanation:
Given that:
Volume (V1) = 2.50 L
Temperature (T1) = 298 K
Volume (V2) = 4.50 L
at standard temperature and pressure;
Pressure (P1) = 1 atm
Temperature (T2) = 273 K
Pressure P2 = ??
Using combined gas law:
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:
Percentage Yield is given as,
%age Yield = Actual Yield / Theoretical Yield × 100
This shows that the %age yield is directly depending upon the actual yield. And most of the time the percentage yield is less than 100 % because of the following factors.
Impure Starting Materials:
If the starting materials (reactants) are not pure then reaction will not completely form the desired product. Different by products will form which will decrease the %age yield.
Incomplete Reactions:
Not all reactions go to completion. In many reactions the starting material after some time stops forming the product due to different conditions. Some reactions attain equilibrium and stop increasing the amount of product. While, in some reactions a by products (like water) formed often react with the product to give a reverse reactions. Hence, the chemistry of reactions also causes the decrease in %age yield.
Handling:
Another major reason for decrease in yield is handling the product. Always some of the product is lost during the workup of the reaction like, taking TLC, doing solvent extraction, doing column chromatography, taking characterization spectrums. So, we can conclude that the %age yield will always be less than 100%.
Answer is: an oxybromate compound is KBrO₄ (x = 4).
ω(Br) = 43.66% ÷ 100%.
ω(Br) = 0.4366; mass percentage of bromine.
If we take 100 grams of compound:
m(Br) = ω(Br) · 100 g.
m(Br) = 0.4366 · 100 g.
m(Br) = 43.66 g; mass of bromine.
n(Br) = m(Br) ÷ M(Br).
n(Br) = 43.66 g ÷ 79.9 g/mol,
n(Br) = 0.55 mol; amoun of bromine.
From chemical formula (KBrOₓ), amount of potassium is equal to amount of bromine: n(Br) = n(K).
m(K) = 0.55 mol · 39.1 g/mol.
m(K) = 21.365 g; mass of potassium in the compound.
m(O) = 100 g - 21.365 g - 43.66 g.
m(O) =34.97 g; mass of oxygen.
n(O) = 34.97 g ÷ 16 g/mol.
n(O) = 2.185 mol.
n(K) : n(Br) : n(O) = 0.55 mol : 0.55 mol : 2.185 mol /÷ 0.55 mol.
n(K) : n(Br) : n(O) = 1 : 1 : 4.
Answer:
The speed of light, c, equals the wavelength, λ (pronounced lambda), times the frequency, ν, (pronounced noo).
c=λν
c is a constant. It is usually given as 3.00×108 m/s or 3.00×1010 cm/s rounded to three significant figures.❤
