A patient who is prescribed a dose inhaler will find that it must be filled with a) medicine in powder form only. Works with lower (not upper) respiratory diseases only. Full of medicine used to give a fixed amount of medicine per oral inhalation. d) Medication in the form of a spray only.
Democritus was the first person to theorize the existence of atoms.
Answer:
The appropriate answer is "9.225 g".
Explanation:
Given:
Required level,
= 63 ppm
Initial concentration,
= 22 ppm
Now,
The amount of free SO₂ will be:
=
=
=
The amount of free SO₂ to be added will be:
=
=
∵ 1000 mg = 1 g
So,
=
=
Thus,
"9.225 g" should be added.
<span>atomic weights: Al = 26.98, Cl = 35.45
In this reaction; 2Al = 53.96 and 3Cl2 = 212.7
Ratio of Al:Cl = 53.96/212.7 = 0.2537 that is approximately four times the mass Cl is needed.
Step 2:
(a) Ratio of Al:Cl = 2.70/4.05 = 0.6667
since the ratio is greater than 0.2537 the divisor which is Cl is not big enough to give a smaller ratio equal to 0.2537.
so Cl is limiting
(b)since Cl is the limiting reactant 4.05g will be used to determine the mass of AlCl3 that can be produced.
From Step 1:
212.7g of Cl will produce 266.66g AlCl3
212.7g = 266.66g
4.05g = x
x = 5.08g of AlCl3 can be produced
(c)
Al:Cl = 0.2537
Al:Cl = Al:4.05 = 0.2537
mass of Al used in reaction = 4.05 x 0.2537 = 1.027g
Excess reactant = 2.70 - 1.027 = 1.67g
King Leo · 9 years ago</span>
The number of formula units in 2.50 mol of the compound is 15.1 * 10^23.
The question is unclear whether NaNO2 or NaNO3 is implied. However,in either case, the solution applies equally.
6.02 * 10^23 formula units of the compound are contained in 1 mole
x formula units are contained in 2.5 moles of the compound
x = 6.02 * 10^23 formula units * 2.5 moles/ 1 mole
x = 15.1 * 10^23 formula units of the compound.
Learn more; brainly.com/question/9743981