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Svetllana [295]
2 years ago
5

What is the measurement 222.008 mm rounded off to four significant digits?

Chemistry
2 answers:
strojnjashka [21]2 years ago
7 0
222.008+eh=345.003 it is that 
Novosadov [1.4K]2 years ago
6 0
That would be 222.0 mm to 4 sig figs

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PLease help me with this question
lukranit [14]
The answer is 6 because the number of chlorine is 2 so if you have 6 moles of chlorine the answer is 6
7 0
3 years ago
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The use of uranium-238 to determine the age of a geological formation is a beneficial use of
tester [92]

Answer:

4- radioactive isotopes

Explanation:

I don't remember exactly but this question was on the regents

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2 years ago
Calculate the percent dissociation of trimethylacetic acid in a aqueous solution of the stuff. You may find some useful data in
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3 0
3 years ago
In acidic solution, the sulfate ion can be used to react with a number of metal ions. One such reaction is SO42−(aq)+Sn2+(aq)→H2
allsm [11]

Answer:

The final balanced equation is :

SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)

Explanation:

SO_4^{2-}(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+Sn^{4+}(aq)

Balancing in acidic medium:

First we will determine the oxidation and reduction reaction from the givne reaction :

Oxidation:

Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)

Balance the charge by adding 2 electrons on product side:

Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)+2e^-....[1]

Reduction :

SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)

Balance O by adding water on required side:

SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)

Now, balance H by adding H^+ on the required side:

SO_4^{2-}(aq)+4H^+(aq)\rightarrow H_2SO_3(aq)+H_2O(l)

At last balance the charge by adding electrons on the side where positive charge is more:

SO_4^{2-}(aq)+4H^+(aq)+2e^-\rightarrow H_2SO_3(aq)+H_2O(l)..[2]

Adding [1] and [2]:

SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)

The final balanced equation is :

SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)

4 0
3 years ago
Select the reason why two molecules of benzaldehyde cannot react via the aldol condensation.
Lesechka [4]
Benzaldehyde or C6H5CHO would not undergo the aldol condensation because it does not contain an alpha-hydrogen in its structure. Aldol condensation is a type of reaction that happens between an enolate and an aldehyde or ketone leading to a alkene that has a planar structure. The lack of an alpha-hydrogen would not allow for it to undergo such process since it cannot enolize. Benzaldehyde undergoes a nucleophilic reaction known as Claisen-Schmidt condensation. It has somehow same mechanism of the aldol reaction however, the nucleophilic attack on the carbonyl happens even without the alpha-hydrogen but with an enolate that is from a ketone.
4 0
3 years ago
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