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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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anyanavicka [17]

Answer:

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Explanation:

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3 years ago
A solution is made by mixing of water and of acetic acid . Calculate the mole fraction of water in this solution. Be sure your a
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Answer:

The answer is "0.35".

Explanation:

please find the complete question in the attached file.

Mass of CH_3C0_2 H= 77 \ g \\\\

Molar mass of CH_3C0_2 H = 60.05 \  \frac{g}{mol} \\\\

No of moles in n_{CH_3CO_2H} = 77 \ g \times  \frac{1 \ mol \ CH_3C0_2H}{60.05 \ g}  

                                          = 1.28 \ mol \ CH_3CO_2H

Mass of water (H_2O)= 42 \ g

The molar mass of water (H_2O)= 18.02 \  \frac{g}{mol}

moles of water n_{H_2O}:

= 42 \ g \times \frac{1 mol H_2O}{18.02 \ g} \\\\= 2.33 \ mol \ H_2 O

Molfraction of acetic acid:

X CH_2CO_2H = \frac{n_{CH_3CO_2H}}{n_CH_3CO_2H +n_{H_2}}\\\\

                     =\frac{1.28 \ mol}{1.28 \ mo1 + 2.33 mol}\\\\ = 0.354\\\\= 0.35

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3 years ago
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