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Ahat [919]
1 year ago
10

question 11 use the internet to look up the sds for 2.0 m sodium hydroxide, naoh, to answer the following questions: a) list the

potential acute and chronic health effects. b) identify the first aid measures for ingestion. c) identify whether or not it is flammable d) identify the chemicals that potentially produce a dangerous reaction with naoh. e) describe how to handle small spills and the personal protective required to work with naoh.
Chemistry
1 answer:
frozen [14]1 year ago
5 0

The compound is basic in nature but the concentration substance is corrosive.

<h3>What is NaOH?</h3>

Sodium hydroxide is a compound that is formed from sodium and hydrogen as well as oxygen The compound is basic in nature but the concentration substance is corrosive.

a) The acute health effect of exposure to NaOH is skin burn while the chronic effect of exposure is damage to the lungs.

b) The first aid for the ingestion of NaOH is to take in plenty of water.

c) The compound is not flammable

d) The chemicals that could produce a dangerous reaction with NaOH are acids

e) To handle small spills, irrigate the affected area with plenty water and use lab coats, gloves and eye wear when working with the solution.

Learn more about sodium hydroxide:brainly.com/question/24010534

#SPJ1

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lilavasa [31]

1.66 M is the concentration of the chemist's working solution.

<h3>What is molarity?</h3>

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.

In this case, we have a solution of Zn(NO₃)₂.

The chemist wants to prepare a dilute solution of this reactant.

The stock solution of the nitrate has a concentration of 4.93 M, and he wants to prepare 620 mL of a more dilute concentration of the same solution. He adds 210 mL of the stock and completes it with water until it reaches 620 mL.

We want to know the concentration of this diluted solution.

As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the diluted solution so:

n_1= n_2  (1)

and we also know that:

n = M x V_2

If we replace this expression in (1) we have:

M_1 x V_1= M_2 x V_2

Where 1, would be the stock solution and 2, the solution we want to prepare.

So, we already know the concentration and volume used of the stock solution and the desired volume of the diluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is M_2:

4.93 x 210 =  620 xM_2

M_2 = 1.66 M

This is the concentration of the solution prepared.

Learn more about molarity here:

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