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1 year ago

use the internet to look up the sds for 2.0 m sodium hydroxide naoh to answer the following question a) list the potential acute

1 answer:

6 0

A safety data sheet is informational data that states the properties of the various chemicals. According to the SDS of NaOH, it causes irritation, nausea, eye burn, burns in the digestive system, etc.

SDS is abbreviated for a safety data sheet which is a document that provides information on the chemicals including the properties, health hazards, measures, safety, prevention, etc.

According to SDS, NaOH results in various potential acute health effects including vomiting, diarrhea, nausea, irritation, etc. The chronic effect includes coma, burns in the eye, respiratory infection, digestive system, dermatitis, etc.

Therefore, sodium hydroxide has various acute and chronic health effects.

Learn more about SDS, here:

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Your question is incomplete, but most probably your full question was, Use the internet to look up the sds for 2.0 m sodium hydroxide, NaOH, to answer the following question: list the potential acute and chronic health effects.

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Calculate the pka of hypochlorous acid. The ph of a 0.015 m solution of hypochlorous acid has a ph of 4.64.

o-na [289]

Answer:

pKa = 7.46

Explanation:

a) Hypochlorous acid = HClO

b) [HClO} = 0.015

c) pH = 4.64

d) pKa = ?

<u>2) Strategy:</u>

With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.

<u>3) Solution:</u>

a) pH

pH = - log [H₃O⁺]

4.64 = - log [H₃O⁺]

$[H_3O^+]= 10^{-4.64} = 2.29.10^{-5}$

b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)

c) Equilibrium constant: Ka = [ClO⁻] [H₃O⁺] / [HClO]

d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M

e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M

f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46

5 0

3 years ago

Calculate the pH after the addition of 10.0 mL of 0.240 M sodium hydroxide to 50.0 mL of 0.120 M acetic acid.

Goryan [66]

Answer:

pH = 4.58

Explanation:

The reaction of NaOH with acetic acid, CH₃COOH occurs as follows:

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

<em>Moles that react:</em>

NaOH = 10mL = 0.010L * (0,240mol / L) = 0.0024 moles NaOH

CH₃COOH = 50.0mL = 0.050L * (0.120mol / L) = 0.0060 moles CH₃COOH

That means after the reaction you will have:

CH₃COOH: 0.0060 mol - 0.0024 mol = 0.0036 moles

CH₃COO⁻Na⁺: 0.0024 moles

in solution, you will have the mixture of a weak acid (Acetic acid), with its conjugate base (sodium acetate, CH₃COO⁻Na⁺). And pH of this buffer can be determined using H-H equation:

pH = pKa + log [A⁻] / [HA]

For Acetic buffer pKa = 4.76:

pH = 4.76 + log [CH₃COO⁻Na⁺] / [CH₃COOH]

<em>Where [] is molarity of each species or moles</em>

<em />

Replacing:

pH = 4.76 + log [0.0024 moles] / [0.0036 moles]

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What is the mass, in grams, of 28.74 mL of acetone?

a_sh-v [17]

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3 years ago

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What is the molar ratio of Li to N2 in the given reaction?<br><br> 6Li + N2 → 2Li3N

givi [52]

Answer:

6moles of Li : 1 mole of N₂

Explanation:

The reaction equation is given as:

6Li + N₂ → 2Li₃N

Mole ratio of Li to N₂;

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6 moles of Li will react with 1 mole of N₂;

The mole ratio is:

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