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1 year ago

What type of cleaning agent should be used to clean a laminar flow hood?

Chemistry

1 answer:

lara31 [8.8K]1 year ago

6 0

The type of cleaning agent that should be used to clean a laminar flow hood is 70% Isopropyl Alcohol.

Isopropyl alcohol is mixed with water for use as a rubbing-alcohol antiseptic. It is also utilized in aftershave creams, hand lotions, and other cosmetics. In enterprise, it is used as a less expensive solvent for cosmetics, drugs, shellacs, and gums, in addition to denaturing ethanol (ethyl alcohol).

Isopropyl alcohol is a colorless, flammable chemical compound with a robust alcoholic smell. As an isopropyl institution related to a hydroxyl group, it's far the handiest example of a secondary alcohol, wherein the alcohol carbon atom is connected to 2 other carbon atoms.

Isopropyl alcohol is pure alcohol and is a colorless liquid with a musty, sharp odor. There are no other components in a bottle of isopropyl alcohol. through assessment, rubbing alcohol consists of isopropyl alcohol among different components, including water. maximum rubbing alcohol manufacturers incorporate 70% isopropyl alcohol.

Learn more about isopropyl alcohol here brainly.com/question/5425836

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Melting water is an example of a physical change because (3 points)

ladessa [460]

The water changes state.

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3 years ago

What is the vapor pressure of an aqueous solution made up of 60.0g of urea (CH4N2O) in 180g of water? the vapor pressure of wate

djyliett [7]

To get the answer you use the Law of Raoult.

Raoult's law states that the decrease of the vapor pressure of a liquid is proportional to the molar fraction of the solute.

ΔP = Pa * Xa

Here Pa = 0.038 atm

And Xa = N a / (Na + Nb), where Na is number of moles of A and Nb is number of moles of b

Na = mass of urea / molar mass of urea = 60 g / (molar mass of CH4N2O)

molar mass of CH4N2O = 12 g/mol + 4*1g/mol + 2*14 g/mol + 16 g/mol = 60 g/mol

Na = 60 g / 60 g/mol = 1 mol

Nb = mass of water / molar mass of water = 180g / 18g/mol = 10 mol

Xa = 1 mol / (10 mol + 1 mol) = 1/11 =0.09091

ΔP = Pb * Xa = 0.038 atm * 0.09091 = 0.0035 atm

Then, the final vapor pressure of water is Pb - ΔP = 0.038atm - 0.0035atm = 0.035 atm.

Answer: 0.035 atm

7 0

3 years ago

(02.02 LC) The two main types of weathering are Select one:

Firlakuza [10]

Answer:

Explanation:

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3 years ago

I need help with my quiz.

katrin2010 [14]

Answer:

Explanation:

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3 years ago

Read 2 more answers

The equilibrium constant has been estimated to be 0.12 at 25 °C. If you had originally placed 0.069 mol of cyclohexane in a 2.8

scZoUnD [109]

Answer: Concentrations of cyclohexane and methylcyclopentane at equilibrium are 0.0223 M and 0.0027 M respectively

Explanation:

Moles of cyclohexane = 0.069 mole

Volume of solution = 2.8 L

Initial concentration of cyclohexane = $\frac{moles}{Volume}=\frac{0.069}{2.8}=0.025M$

The given balanced equilibrium reaction is,

cyclohexane ⇔ methylcyclopentane

Initial conc. 0.025 M 0

At eqm. conc. (0.025-x)M (x) M

The expression for equilibrium constant for this reaction will be,

K= methylcyclopentane / cyclohexane

Now put all the given values in this expression, we get :

$0.12=\frac{(x)}{(0.025-x)}$

By solving the term 'x', we get :

x = 0.0027

Concentration of cyclohexane at equilibrium = (0.025-x ) M = (0.025-0.0027) M = 0.0223 M

Concentration of methylcyclopentane at equilibrium = (x ) M = (0.0027) M

4 0

3 years ago