All categories

4 years ago

What is budesonide inhalation suspension?

1 answer:

6 0

GENERIC NAME: BUDESONIDE SUSPENSION FOR NEBULIZER - INHALATION (bue-DES-oh-nide) ... USES: Budesonide is used to control and prevent symptoms (wheezing and shortness of breath) caused by asthma. This medication belongs to a class of drugs known as corticosteroids.

You might be interested in

Prove the following:V=U + AT

pychu [463]

We know that,

acceleration (a) = final velocity(V)-initial velocity(U)/ Time taken(T)

6 0

2 years ago

The relatively high boiling point of water is due to water having

Elden [556K]

The answer is A because water molecules don't have metallic bonding, and water is also polar, and it takes more time for the heat energy to break down the hydrogen bonds

7 0

3 years ago

Which of these is an example of a chemical change?

dolphi86 [110]

The answer to this is D.

3 0

4 years ago

Read 2 more answers

Determine which molecules are weak acids and weak bases, and place them in the appropriate category.

Dmitry [639]

Answer:

$CH_{3} COOH$ is a weak acid with a strong conjugate base $(CH_{3} CO_{2} ^{-} )$

$CH_{3} NH_{2}$ is a weak base with a strong conjugate acid $(CHx_{3} NH_{2} ^{-} )$

Explanation:

Weak acids do not completely dissociate in water to its distinctive ions, but instead forms conjugate bases; and ice versa with weak bases. The conjugate base has strong affinity for protons and vice versa.

I hope you found this explanation clear and easy to follow.

3 0

3 years ago

Ozone, o3, is a product in automobile exhaust by the reaction represented by the equation no2(g) + o2(g) --> 3no(g) + o3(g).

svetoff [14.1K]

Answer: 2.1 g mass of ozone( $O_{3}$ ) is predicted to form from the reaction of 2.0 g $NO_{2}$ in a car's exhaust and excess oxygen

Given information : Mass of $NO_{2}$ = 2.0 g and $O_{2}$ is in excess.

We need to calculate the mass of ozone ( $O_{3}$ )

Mass of ozone( $O_{3}$ ) is calculated with the help of mass of $NO_{2}$ using stoichiometry.

$NO_{2} + O_{2}\rightarrow NO + O_{3}$

Step 1 : Convert grams of $NO_{2}$ to moles of $NO_{2}$ .

$Moles = \frac{Grams}{Molar mass}$

Molar mass of $NO_{2}$ = 46.0 g/mol

$Moles = \frac{2.0g}{46.0\frac{g}{mol}}$

Moles of $NO_{2}$ = 0.043 mol

Step 2 : Find the moles of $O_{3}$ using moles of $NO_{2}$ .

Moles of $O_{3}$ is calculated by using moles of $NO_{2}$ with the help of mole ratio.

A mole ratio is the ratio between the amounts in moles of any two compounds involved in a chemical reaction. The mole ratio may be determined by examining the coefficients in front of formulas in a balanced chemical equation.

From the balanced chemical equation we can see that coefficient of $NO_{2}$ is 1 and coefficient of O3 is 1 , so mole ratio of $O_{3}$ to $NO_{2}$ is 1:1

Moles of $O_{3}$ = $(0.043 mol NO_{2})\times \frac{(1 mol O_{3})}{(1 mol NO_{2})}$

Moles of $O_{3}$ = $(0.043)\times \frac{(1 mol O_{3})}{(1)}$

Moles of $O_{3}$ = 0.043 mol

Step 3 : Convert moles of $O_{3}$ to grams of $O_{3}$

Grams = Moles X Molar mass

Molar mass of $O_{3}$ = 48.0 g/mol

Grams = $(0.043 mol O_{3})\times (\frac{48 g O_{3}}{1 mol O_{3}})$

Grams = $(0.043)\times (\frac{48 g O_{3}}{1})$

Grams = 2.1 g $O_{3}$

Note : The above three steps can also be done using a single step setup.

Grams of $O_{3}$ = $(2.0 gNO_{2})\times \frac{(1mol NO_{2})}{(46.0 g NO_{2})}\times \frac{(1 mol O_{3})}{(1 mol NO_{2})}\times \frac{(48.0 g O_{3})}{(1 mol O_{3})}$

Grams of $O_{3}$ = $(2.0 )\times \frac{(1)}{(46.0 )}\times \frac{(1)}{(1)}\times \frac{(48.0 g O_{3})}{(1)}$

Grams of $O_{3}$ = 2.1 grams

4 0

3 years ago