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Alexandra [31]
2 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]2 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:

no

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5 0
2 years ago
What is the net cell reaction for the iron-silver voltaic cell?
RSB [31]
The net cell reaction for the iron-silver voltaic cell would be:

<span>3Ag+ + Fe --------> 3Ag + Fe3+
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wherein it comes from the reaction of the cathode and the anode:

<span>Ag+ + e- -------> Ag, for the cathode and;
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<span>Fe -------> Fe3+ + 3e-, for the anode
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When both equations would be multiplied, the it would now yield the balanced net reaction as stated above.
4 0
2 years ago
Given the connection between Aw and K (Aw=2k) could you use the ideal gas law and derive the Boltzmann constant. Water freezes a
Elena L [17]

Answer:

Explanation:

The relation between new scale and absolute temperature scale is given as follows

Aw = 2 K

for K = 273.15 ( freezing point of water at absolute scale )

Aw = 2 x 273.15 = 546.3 K

So each division of new scale is half the each division of absolute scale

each division of new scale is small .

The value of R  = 8.314 J per mole per K

Here per K is equivalent to 2Aw

So the vale of R in new scale = 8.314/2 J per mole per Aw

=  4.157 J per mole per Aw

k = R / N

= 4.157 / 6.02 x 10²³

= .69 x 10⁻²³

= 6.9 x 10⁻²⁴ J per molecule per Aw .

7 0
2 years ago
Read 2 more answers
Calculate the change in entropy that occurs in the system when 2.70 mole of diethyl ether (C4H6O) condenses from a gas to a liqu
AlladinOne [14]

Answer:233 Joules/K

Explanation:

∆H= 26.5KJmol-1

Kelvin temperature = 34.6 + 273 = 307.6 K

No of moles= 2.7 moles

2.70 mole x 26.5 kJ/mole = 71.55kilojoules

∆S=71.55 kilojoules / 307.6 K = 0.233 kilojoules/ K

Convert to JK-1

0.233 kilojoules/ K x 1000 Joules/kilojoule = 233 Joules/K

5 0
2 years ago
Calculate the ph of a solution formed by mixing 150.0 ml of 0.10 m hc7h5o2 with 100.0 ml of 0.30 m nac7h5o2. The ka for hc7h5o2
m_a_m_a [10]

Answer:

4.49

Explanation:

pKa = - log 6.5 x 10⁻⁵

pKa  =4.19

Given that :

Volume = 150 mL = 0.150 L

For solutions:

number of moles of acid = volume × concentration

number of moles of  acid = 0.150 L × 0.10 M = 0.0150

number of moles  of salt = 0.100 L × 0.30 M = 0.0300

total volume = 150 + 100

= 250 mL

= 0.250 L

Concentration = number of moles/ Volume

∴

For [salt] = 0.0300/ 0.250

= 0.12 M

For [acid] = 0.0150/ 0.250

=0.06 M

pH = 4.19 + log 0.12/0.06

=4.49

7 0
2 years ago
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