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Alexandra [31]
3 years ago
13

Which would be the best to neutralize a large acid spill in your school lab: sodium hydroxide or baking soda? Explain.

Chemistry
1 answer:
nadya68 [22]3 years ago
5 0

Consider the acid spill. It is already starting to do nasty things to, say, the floor or counter. So you grab the bottle of 10% NaOH and pour some on the spill. All of a sudden, you get a great deal of heat, and you don't have any visual evidence whether your put on too little or too much. But you have added more liquid to the spill, generated more heat, and will get more damage. You have made a bigger mess, and if you added too much, you then have a neutralization problem to deal with.  

And if it is something like a strong sulfuric acid solution, adding sodium hydroxide solution will be extremely exothermic, and you could get some really nasty results.  

So now approach the spill with a handful of baking soda. You sprinkle it on the spill. It fizzes, and carbon dioxide is given off. That actually, in a very tiny way, moderates the temperature of the neutralization. And you can keep adding baking soda until the fizzing stops, and then perhaps some water to mix everything well. But what you have done is kept the volume to a minimum, added a neutralization agent that has a visible endpoint (no more gas being given off), and you don't suddenly have a huge amount of highly basic solution because you added too much.  

And what is also nice about baking soda is that you can toss some with your hand or even with a spoon, and get some distance from the spill. With a liquid, you have to get much closer

i hope this helped..

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Answer:

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3 years ago
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When 32 grams of aluminum react, the actual yield is 105.5 grams, what is the percent yield?
user100 [1]

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Explanation:

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3 years ago
A sample of methane gas, CH4, occupies 3.25 L at temperature of 19.0 o C. If the pressure is held constant, what will be the tem
Artemon [7]

Answer:

625.46 °C

Explanation:

We'll begin by converting 19 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

T(°C) = 19 °C

T(K) = 19 °C + 273

T(K) = 292 K

Next, we shall determine the Final temperature. This can be obtained as follow:

Initial volume (V₁) = 3.25 L

Initial temperature (T₁) = 292 K

Final volume (V₂) = 10 L

Final temperature (T₂) =?

V₁/T₁ = V₂/T₂

3.25 / 292 = 10 / T₂

Cross multiply

3.25 × T₂ = 292 × 10

3.25 × T₂ = 2920

Divide both side by 3.25

T₂ = 2920 / 3.25

T₂ = 898.46 K

Finally, we shall convert 898.46 K to celsius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

T(K) = 898.46 K

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4 0
3 years ago
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The chemical formula : 3HgBr₂(Mercury(II) bromide)

<h3>Further explanation</h3>

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Required

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The number of molecules is determined by the coefficient in front of the compound

the number of atoms is determined by the subscript after the atom and the coefficient

Three molecules⇒ coefficient = 3

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The chemical formula : 3HgBr₂

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