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Zigmanuir [339]
2 years ago
13

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

Chemistry
1 answer:
VashaNatasha [74]2 years ago
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.

<h3>What is SDS?</h3>

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:

brainly.com/question/2471127

#SPJ4

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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This results in higher temperature and pressure which in turn causes the outer shells to expand and cool as a red giant.

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3 years ago
The pKa for a weak acid to be used in the buffer should be as close as possible to the desired pH. What if the pKa is not close
nikdorinn [45]

Answer:

When pH = pKa, the ionizable compound in question (either acid or base) will be half protonated and half deprotonated

Explanation:

A convenient way of expressing the relative strength of an acid is by the value of its pKa, which makes it easy to see in small changes in pKa the changes associated with large variations in Ka. Small pKa values are equivalent to large Ka (dissociation constant) values, and as pKa decreases, the strength of the acid increases.

An acid will be stronger the lower its pKa and in a base it happens the other way around, which is stronger the higher its pKa.

Those dissociation constants are not fixed, they depend on other variables. For example, the dissociation constant changes at different temperatures. However, it maintains its value at the same temperature, before changes in the concentration of any of the species or even under the action of a catalyst.

3 0
4 years ago
Which of the pairs of compounds below could be used to prepare a buffer solution? hcl and nacl hc2h3o2 and kc2h3o2 nh3 and nh4no
Basile [38]
We have that the correct answer is the last one. NH4NO3 is called ammonium nitrate and is the salt that appears from combining HNO3 (acid) with NH3 (ammonia, base). We have then that a solution of this type is a buffer solution because ammonia is a weak base and the salt contains its conjugate acid (definition of buffer solution). The 2nd example is wrong because both potassium acetate and Acetic Acid are acids; no conjugate base is involved. HCI and NaCl are pairs of conjugacy (HCI is an acid and NACI can be produced by NaOH and HCI), but HCI is a strong acid and NACI is a salt. Hence, it cannot be a buffer solution (weak acid needed). Weak acid or base means that it does not fully break down in ions when in water.
8 0
4 years ago
The oxidation of ammonia produces nitrogen and water via the following reaction: 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l) Suppose the
Sonbull [250]

Answer:

The rate of consumption of NH_{3} is 2.0 mol/L.s

Explanation:

Applying law of mass action to this reaction-

-\frac{1}{4}\frac{\Delta [NH_{3}]}{\Delta t}=-\frac{1}{3}\frac{\Delta [O_{2}]}{\Delta t}=\frac{1}{2}\frac{\Delta [N_{2}]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_{2}O]}{\Delta t}

where -\frac{\Delta [NH_{3}]}{\Delta t} represents rate of consumption of NH_{3}, -\frac{\Delta [O_{2}]}{\Delta t} represents rate of consumption of O_{2}, \frac{\Delta [N_{2}]}{\Delta t} represents rate of formation of N_{2} and \frac{\Delta [H_{2}O]}{\Delta t} represents rate of formation of H_{2}O.

Here rate of formation of H_{2}O is 3.0 mol/(L.s)

From the above equation we can write-

-\frac{1}{4}\frac{\Delta [NH_{3}]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_{2}O]}{\Delta t}

Here \frac{\Delta [H_{2}O]}{\Delta t}=3.0 mol/(L.s))

So, -\frac{\Delta [NH_{3}]}{\Delta t}=\frac{4}{6}\frac{\Delta [H_{2}O]}{\Delta t}

Hence, -\frac{\Delta [NH_{3}]}{\Delta t}=\frac{4}{6}\times 3.0 mol/(L.s)=2.0 mol/(L.s)  

6 0
3 years ago
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