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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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A sample of 25mls of HCL was mixed with 25mls of KOH of the same concentration in a calorimeter. As a result the temperature ros

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2 years ago

How many moles of NaCl are present in 2.50 L of a 0.070 M solution?

posledela

Given :

Volume of NaCl solution 2.5 L .

Molarity of NaCl solution is 0.070 M .

To Find :

How many moles are present in the solution.

Solution :

Let, n be the number of moles.

We know, molarity is given by :

$M = \dfrac{n}{V(in\ L)}$

So,

$n = M \times V\\\\n = 0.070\times 2.5 \\\\n = 0.175\ moles$

Therefore, number of moles of NaCl is 0.175 moles.

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2 years ago

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Aleks04 [339]

Answer:

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2 years ago

Read 2 more answers

when a solution of sodium chloride is added to a solution of copper(ii) nitrate, no precipitate is observed. Write the molexular

zhuklara [117]

Explanation:

Cu(NO3)2 + 2NaCl(aq) --> CuCl2(aq) + 2NaNO3(aq)

Cu(NO3)2 + 2NaOH(aq) --> Cu(OH)2(s) + 2NaNO3(aq)

A light blue precipitate of Cu(OH)2 is formed and NaNO3 in solution.

Cu(NO3)2(aq) --> Cu2+(aq) + 2NO3^-2(aq)

2NaOH(aq) --> 2Na+(aq) + 2OH-(aq)

Cu2+(aq) + 2OH-(aq) --> Cu(OH)2(aq)

2Na+(aq) + 2NO3^-2(aq) --> 2NaNO3(aq)

The reaction in both Questions 1 and 2 is called Double displacement reaction. A double-replacement reaction exchanges the cations and/or or the anions of two ionic compounds. A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate (precipitated) while the other in solution.

Since the cation and anions in Qustion 1 were exchanged, the same was done for Question 2, hence the identity of the precipitate in Question 2 was got.

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2 years ago

Doping Se with P would produce a(n) ________ semiconductor with ____________ conductivity compared to pure Se.

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Doping Se (group VI elements) with P(group V)elements would produce a P-TYPE semiconductor with HIGHER conductivity compared to pure Se

the reason is P dopant will introduce holes in the Se as P has lesser valence electron

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2 years ago