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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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Complete the table by classifying each property as either a physical or chemical property
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If an atom has an atomic number of 6 and and abdomens mass of 12 how many electrons does it have
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5 0
3 years ago
Use the formation reactions below such that when added together, they match the balanced equation for the combustion of methane.
muminat

Answer:

ΔH of the reaction is -802.3kJ.

Explanation:

Using Hess's law, you can know ΔH of reaction by the sum of ΔH's of half-reactions.

Using the reactions:

<em>(1) </em>Cgraphite(s)+ 2H₂(g) → CH₄(g) ΔH₁ = −74.80kJ

<em>(2) </em>Cgraphite(s)+ O₂(g) → CO₂(g) ΔH₂ = −393.5k J

<em>(3) </em>H₂(g) + 1/2 O₂(g) → H₂O(g) ΔH₃ = −241.80kJ

The sum of (2) - (1) produce:

CH₄(g) + O₂(g) → CO₂(g) + 2H₂(g) ΔH' = -393.5kJ - (-74.80kJ) = -318.7kJ

And the sum of this reaction with 2×(3) produce:

CH₄(g) + 2 O₂(g) → CO₂(g) + 2H₂O(g) And ΔH = -318.7kJ + 2×(-241.80kJ) =

<em>-802.3kJ</em>

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3 years ago
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