What is the HCPCS Level II code for the compounded concentrated form of 0.5 mg Levalbuterol HCL when inhaled

Answers

Answer 1

Answer:

aefhfcjd

Explanation:


Related Questions

cuantas moléculas de oxigeno se producen por la descomposición de 28.5 g de H2O2 (masa molecular = 34.0g/mol) de acuerdo a la ecuación
2H2O2(l) → 2H2O(l)+O2(g)

Answers

The question is as follows: How many oxygen molecules are produced by the decomposition of 28.5 g of H2O2 (molecular mass = 34.0g / mol) according to the equation

2H2O2 (l) → 2H2O (l) + O2 (g)

Answer: There are [tex]2.52 \times 10^{23}[/tex] molecules are produced by the decomposition of 28.5 g of [tex]H_{2}O_{2}[/tex] according to the equation [tex]2H_{2}O(l) \rightarrow 2H_{2}O(l) + O_{2}(g)[/tex].

Explanation:

Given: Mass of [tex]H_{2}O_{2}[/tex] = 28.5 g

As moles is the mass of a substance divided by its molar mass. Hence, moles of [tex]H_{2}O_{2}[/tex] is calculated as follow.

[tex]Moles = \frac{mass}{molarmass}\\= \frac{28.5 g}{34.0 g/mol}\\= 0.838 mol[/tex]

According to the given equation, 2 moles of [tex]H_{2}O_{2}[/tex] gives 1 mole of [tex]O_{2}[/tex]. So, moles of [tex]O_{2}[/tex] produced by 0.838 moles of [tex]H_{2}O_{2}[/tex]  will be calculated as follows.

[tex]Moles of O_{2} = \frac{0.838 mol}{2}\\= 0.419 mol[/tex]

This means that moles of [tex]O_{2}[/tex] produced is 0.419 mol.

As per the mole concept, 1 mole of every substance has [tex]6.022 \times 10^{23}[/tex] molecules.

So, molecules of [tex]O_{2}[/tex] present in 0.419 mole are as follows.

[tex]0.419 \times 6.022 \times 10^{23}\\= 2.52 \times 10^{23}[/tex]

Thus, we can conclude that there are [tex]2.52 \times 10^{23}[/tex] molecules are produced by the decomposition of 28.5 g of [tex]H_{2}O_{2}[/tex] according to the equation [tex]2H_{2}O(l) \rightarrow 2H_{2}O(l) + O_{2}(g)[/tex].

A gas has density 2.41 g/liter at 25°C and 770 mm Hg. Calculate it's molecular mass (R = 0.0821 L atm.mol-1K-1.

ASAP!!!!!!!!! ​

Answers

Answer:

Molecular mass, M = 58.20 g/mol.

Explanation:

Given the following data;

Density = 2.41 g/literTemperature = 25°CPressure = 770 mmHgGas constant, R = 0.0821 L atm.mol-1K-1

Conversion:

760 mmHg = 1 atm

770 mmHg = 770/760 = 1.0131 atm

Temperature = 25°C = 273 + 25 = 298 K

To find the molecular mass, we would use the ideal gas law equation (density version);

PM = dRT

Where;

P is the pressure.M is the molecular mass.d is the density of a substance.R is the ideal gas constant.T is the temperature.

Making M the subject of formula, we have;

[tex] M = \frac {dRT}{P} [/tex]

Substituting into the formula, we have;

[tex] M = \frac {2.41 * 0.0821 * 298}{1.0131} [/tex]

[tex] M = \frac {58.9626}{1.0131} [/tex]

Molecular mass, M = 58.20 g/mol.

True or false
If an atom is charged positive, it contains more protons than electrons.

Answers

Answer:

TRUE

Explanation: because it is positively charged and protons are positive ions

an experiment was carried out and H2SO4 was collected resulting in a 92.0% yield. What mass of H2SO4 was collected from the experiment?

Answers

Answer:

The mass of H2SO4 collected is 92.0g assuming the theoretical yield of H2SO4 is 100g

Explanation:

Assuming the theoretical yield of the sulfuric acid, H2SO4, is 100g.

To solve this question we must use the equation of percent yield:

Percent yield = Actual yield / Theoretical yield * 100

Where percent yield = 92.0%, actual yield is the mass of H2SO4 produced and theoretical yield is 100g, the mass we are assuming

Replacing:

92.0% = Actual yield / 100g * 100

92.0g = Actual Yield

The mass of H2SO4 collected is 92.0g assuming the theoretical yield of H2SO4 is 100g

Na3N decomposes to form sodium and nitrogen gas at STP. If 13.7 L of nitrogen is produced
how many moles of Na3N was used? (22.4 L = 1 mole of any gas)
2Na3N --> 6Na + N2

Answers

Answer: 1.224 moles of [tex]Na_3N[/tex] were used.

Explanation:

We are given:

Volume of nitrogen gas produced = 13.7 L

At STP conditions:

22.4 L of volume is occupied by 1 mole of a gas

Applying unitary method:

13.7 L of nitrogen gas will be occupied by = [tex]\frac{1mol}{22.4L}\times 13.7L=0.612mol[/tex]

For the given chemical reaction:

[tex]2Na_3N\rightarrow 6Na+N_2[/tex]

By Stoichiometry of the reaction:

1 mole of nitrogen gas is produced by 2 mole of [tex]Na_3N[/tex]

So, 0.612 moles of nitrogen gas will be produced from = [tex]\frac{2}{1}\times 0.612=1.224mol[/tex] of [tex]Na_3N[/tex]

Hence, 1.224 moles of [tex]Na_3N[/tex] were used.

explain why the boiling point of aluminum is higher than sodium​

Answers

Explanation:

this is the reason why aluminium's boiling point is higher than that of sodium. sodium forms Na+ ions, therefore 1 electron is lost from each sodium atom, whereas aluminium forms Al3+ ions, so 3 electrons are lost for each Aluminum atom. The number of delocalised electrons therefore increases from sodium to aluminium, and charge density of the ions increases as the ionic charge increases and size decreases.

Why is it essential for a calorimetry to be an insulated (closed) system?


A. The heat exchange must originate from only the reaction (system).

B. Thermal insulation minimizes energy loses to the atmosphere.

C. A and B

D. None of the Above

Answers

Answer:

B.Thermal insulation minimizes energy loses to the atmosphere.

Explanation:

It is important because it helps to stop hit from transferring from the calorimeter to the environment. This would help to have an accurate measurement of the heat that was used in the chemical process. The greatest cause of error that happens in calorimetry is when heat is lost to the environment. To reduce this, you insulate the calorimeter and add a cover.

Which compound is a glycol?
CH3―CH2―CH2―CHO
CH3—CH2—O—CH2—CH3
CH3—CH2—CH2—CH2OH
HOH2C―CH2―CH2―CH2OH

Answers

Answer:

CH3—CH2—O—CH2—CH3 is the compound of glycol.

Answer:

HOH2C―CH2―CH2―CH2OH is glycol

Explanation:

Select the correct text in the passage.

Answers

Explanation:

Oceans can influence weather and climate. For example, cities located near large water bodies like oceans tend not to have extreme climates

Balance equation for. _Mg + _H3(PO4) --_Mg3(PO4)2+ _H2

Answers

3Mg + 2H3(PO4) —> Mg3(PO4)2 + 3H2

I think

Zn-64 = 48.63%
Zn-66 = 27.90%
Zn-67 = 4.10%
Zn-68 = 18.75%
Zn-70 = .62%
Calculate the atomic mass/given their percent abundance

Answers

Answer:

A = 65.46 u

Explanation:

Given that,

The composition of zinc is as follows :

Zn-64 = 48.63%

Zn-66 = 27.90%

Zn-67 = 4.10%

Zn-68 = 18.75%

Zn-70 = .62%

We need to find the  average atomic mass of the given element. It can be solved as follows :

[tex]A=\dfrac{48.63\times 64+27.90\times 66+4.1\times 67+18.75\times 68+0.62\times 70}{100}\\A=65.46\ u[/tex]

So, the average atomic mass of zinc is 65.46 u.

Which expression is equal to f(x) · g(x)?

Answers

Answer:

C. x⁴ + 6·x³ - 12·x  - 72

Explanation:

The given functions are;

[tex]f(x) =\sqrt{x^2 + 12 \cdot x + 36}[/tex]

g(x) = x³ -12

We have that [tex]f(x) =\sqrt{x^2 + 12 \cdot x + 36}[/tex] = [tex]f(x) =\sqrt{(x + 6)^2}[/tex] =  (x + 6)

Therefore;

f(x)·g(x) = [tex]\sqrt{x^2 + 12 \cdot x + 36}[/tex] × (x³ - 12) = (x + 6) × (x³ - 12)

(x + 6) × (x³ - 12) = x⁴ - 12·x + 6·x³ - 72 = x⁴ + 6·x³ - 12·x  - 72

∴ f(x)·g(x) = [tex]\sqrt{x^2 + 12 \cdot x + 36}[/tex] × (x³ - 12) = x⁴ + 6·x³ - 12·x  - 72

draw the flow chart of production of silk from silk moth

please help....
no links or get reported ​

Answers

Too lazy to draw sorry lol.

#CarryOnLearning

I actually just looked it up via image search.

just copy by hand either of the flow charts.

if you can, use colors to fill. makes it more interesting to look at

and define a circle (or rather a few points on it) to draw it neatly. the defined points can later be transformed into the arrows.

Fe-54 = 5.845%
Fe-56 = 91.754%
Fe-58 = 2.119%
Calculate the average atomic mass/given their percent abundance

Answers

Answer:

[tex]\boxed {\boxed {\sf 55.76756 \ amu}}[/tex]

Explanation:

The average atomic mass is the sum of the product of the given masses and their abundances. We have to complete two steps for each isotope, then add all the products together.

Fe-54

Convert the percent abundance to a decimal by dividing by 100.

5.845/100= 0.05845

Multiply the abundance as a decimal by the mass number. This is the number after the element in the isotope symbol (54 amu).

0.005845 * 54=3.1563 amu

Fe-56

Convert the percent to a decimal.

91.754/100=0.91754

Multiply the decimal by the mass number (56 amu).

0.91754 * 56= 51.38224 amu

Fe-58

Convert the percent to a decimal.

2.119/100=0.02119

Multiply by the mass number (58 amu).

0.02119 * 58 = 1.22902 amu

Average Atomic Mass

Add all the products together to find the average atomic mass.

3.1563 amu + 51.38224 amu + 1.22902 amu =55.76756 amu

The average atomic mass given the percent abundance is 55.7656 atomic mass units.

what is average velocity with formula?

Answers

Answer:

total displacement \time

Explanation:

Dominic needs some bleach to clean up a spill. He finds a bottle of cleaning solution, but there is no label on it. What should he do

Answers

Answer:

He should ask an adult if they know what the solution is. if they dont, put it back and find a different cleaning solution to use

10:38 Fri 9 Jul
ES
GCSE Science - Chemistry
2 of 15
Name the indicator that is red in acid and turns green when the solution becomes neutral.
|

Answers

Answer.

Universal indicator

Explanation.

Universal indicator has many different colour changes, from red for strongly acidic solutions to dark purple for strongly alkaline solutions. In the middle, neutral pH 7 is indicated by green.

True or false, If an atom is charged negative, it contains more electrons than protons.

Answers

False, it would contain more protons than electrons because of the negative charge

Answer:

True

Explanation:

An atom would carry more protons if positively charged, an equal amount of both protons and electrons if neutral, and more electrons if charged negative.

A student is asked to seperate two liquids. Liquid A boils at 100°c and liquid B boils at 65°c. The student sets up a fractional distillation experiment, and after a few minutes a clear liquid is collected from the condenser. Explain which of the two liquids will be collected first. ( 2 marks). ​

Answers

Answer:

B

Explanation:

because B has a lower bp it needs less time and energy to turn into vapour and is collected into the condenser first

A student is asked to separate two liquids. Liquid A boils at 100°C
and liquid B boils at 65 °C. The student sets up a fractional distillation
experiment, and after a few minutes a clear liquid is collected from the
condenser. Explain which of the two liquids will be collected first.

Answers

Answer:

liquid  B

Explanation:

because B has a lower bp it needs less time and energy to turn into vapour and is collected into the condenser first

In the hydrogenation of double bonds, a catalyst is needed. In the third step, the reactants react to form the product. This step is known as

Answers

Answer:

reaction

Explanation:

A reaction is defined as a process where the chemical transformation takes place from one form to some other form. Here, two different chemical substances are reacted together to form some other completely different product.

A hydrogenation process is a chemical process where between the molecular hydrogen and some another element or a compound. This process is usually carried out in the presence of catalyst mainly nickel, platinum or palladium.

Thus, in hydrogenation of a double bond, the catalyst is used to speed up the reaction. The first step in this process is adsorption process, the second step is the diffusion process, the third step is the reaction process and the last step is the desorption process.

A 4.369 g sample of metal is placed in a flask. Water is added to the flask and the total volume in the flask is read to be 126.4 ml. The mass of the water, flask, and metal is 268.5 g. If the mass of the flask is 139.3 g and the density of water is 1.000 g/mL, the density of the solid is ________ g/cm3.

Answers

Answer:

Density of the solid=[tex]2.78 g/cm^3[/tex]

Explanation:

We are given that

Mass of sample of metal=4.369 g

Volume in the flask, V=126.4 ml

Mass of water, flask, and metal=268.5 g

Mass of flask=139.3 g

Density of water=1.000 g/mL

We have to find the density of the solid.

Mass of water=268.5-4.369-139.3=124.831 g

Volume of water=[tex]\frac{Mass\;of\;water}{density\;of\;water}[/tex]

Volume of water=[tex]\frac{124.831}{1}=124.831 mL[/tex]

Volume of solid=126.4 ml-124.831 mL

=1.569mL

Now,

Density of the solid=[tex]\frac{mass\;of\;solid}{volume\;of\;solid}[/tex]

=[tex]\frac{4.369}{1.569}[/tex]

[tex]=2.78g/mL[/tex]

1mL=1 cubic cm

Therefore,

Density of the solid=[tex]2.78 g/cm^3[/tex]

Select the correct answer.
Which missing item would complete this beta decay reaction?

ОА. 0,-1В
OB. 0,0y
OC. 4,2He
OD. 0,1n
Reset

Answers

Answer:

Option A. ⁰₋₁β

Explanation:

Let the unknown be ʸₓA

Thus, the equation given becomes:

⁹⁸₄₃Tc —> ⁹⁸₄₄Ru + ʸₓA

Next, we shall determine the value of x, y and A in order to obtain the answer to the question. This can be obtained as follow:

43 = 44 + x

Collect like terms

43 – 44 =

–1 = x

x = –1

98 = 98 + y

Collect like terms

98 – 98 = y

0 = y

y = 0

ʸₓA => ⁰₋₁A => ⁰₋₁β

Thus, the complete equation is

⁹⁸₄₃Tc —> ⁹⁸₄₄Ru + ⁰₋₁β

The missing item is ⁰₋₁β

Which drawing is structural model of C3H8?

Answers

Answer:

option B is the correct answer

what are neutral salts​

Answers

Answer:

Salts that produce acidic solutions are acid salts. Neutral salts are those salts that are neither acidic nor basic. Zwitterions contain an anionic and a cationic centre in the same molecule, but are not considered to be salts. Examples of zwitterions include amino acids, many metabolites, peptides, and proteins.

Explanation:

Also found in: Thesaurus, Medical, Encyclopedia. (Chem.) a salt formed by the complete replacement of the hydrogen in an acid or base; in the former case by a positive or basic, in the latter by a negative or acid, element or radical.

Determine the excluded volume per mole and the volume actually occupied by a mole for a gas consisting of molecules with radius 167 pm. [Note: To obtain the volume in liters, we must express the radius in decimeters (dm).] Enter your answers in scientific notation.

Answers

Answer:

Explanation:

Volume of one mole of a gas = 22.4 litre.

No of molecules in one mole of gas = 6.02 x 10²³ .

Volume of one molecule = 4/3 π R³

= 4/3 x 3.14 x ( 167 x 10⁻¹² m )³

= 19.5 x 10⁶ x 10⁻³⁶ m³

=19.5 x 10⁻³⁰ m³

= 19.5 x 10⁻³⁰ x 10³ litre .

=  19.5 x 10⁻²⁷ litre .

So volume occupied by molecules in one mole of gas

= 19.5 x 10⁻²⁷ x 6.02 x 10²³  litre

117.4 x 10⁻⁴ litre

= .01174 litre.

Excluded volume

= ( 22.4 -  .01174  ) litre .

= 22.388 litre.

I am having trouble converting :(

Answers

Answer:

For the first question, to determine the total number of molecules of nitrogen dioxide, first make use of the molar mass of the nonpolar compound and then use that to find the total number of moles and then subsequently after make use of the ratio for the Avogadro's number to determine the total number of molecules of this compound.

For the final question, do the inverse, where we make use of the molecules of the compound and then use Avogadro's number to determine the moles of the compound and then use the same molar mass of the compound to determine the grams of the Nitrogen Dioxide.

In an atomic model that includes a nucleus, positive charge is
a
concentrated at multiple sites in an atom.
b
concentrated in the center of an atom.
c
spread evenly throughout an atom.
d
located in the space outside the nucleus.

Answers

The answer is B:concentrated in the center of an atom. Hope this helps.

Which of the following elements has three valence electrons? (2 points)

Li
C
Al
Mg

Answers

Answer:

Al

Explanation:

It has 3.

Trends in the periodic table indicate that the element with the greatest ionization energy is in which of the following periods and groups?

a
Period 2, Group 1
b
Period 7, Group 2
c
Period 6, Group 17
d
Period 1, Group 18

Answers

Answer:

D Period 1, Group 18

Explanation:

it's helium

it's top right corner of the periodic table

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