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GREYUIT [131]
3 years ago
7

Drag the tiles to the correct boxes to complete the pairs.

Chemistry
2 answers:
3241004551 [841]3 years ago
5 0

Answer:

Explanation:

here's the answer hope it helps! :)

7nadin3 [17]3 years ago
3 0

The correct answer is B.

Have a great day!

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QUICK QUESTION: On the Bohr model, how come potassium has 19 electrons in its valence shell if potassium has a K+? Isn’t it supp
Vlada [557]

Answer:  K only has 1 valence electron.  It will leave with only a little effort, leaving behind a positively charged K^+1 atom.

Explanation:  A neutral potassium atom has 19 total electrons.  But only 1 of them is in potassium's valence shell.  Valence shell means the outermost s and p orbitals.  Potasium's electron configuration is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1.  The 4s orbital is the only orbital in the 4th energy level.  So it has a valency of 1.  This means this electron will be the most likely to leave, since it is the lone electron in the oyutermost energy level (4).  When that electron leaves, the charge on the atom go up by 1.  The atom now has a full valence shell of 3s^2 3p^6, the same as argon, Ar.

4 0
2 years ago
What is a sentence for weathering
bija089 [108]
These rocks were either roasted or exposed to severe weathering 
7 0
3 years ago
Read 2 more answers
If hydrochloric acid is obtained commercially at a concentration of 12.1M, how many milliliters of 12.1M HCl(aq) must be used to
rewona [7]

Answer:

V_1=82.6mL

Explanation:

Hello there!

In this case, according to this question, we will need to deal with this dilution problem, because it is asking for the volume of a 12.1-M stock solution of HCl. In such a way, we can use the following equation, under the assumption of no change in the number of moles in the solution:

M_2V_2=M_1V_1

Thus, we solve for the initial volume, V1, as shown below:

V_1=\frac{M_2V_2}{M_1}

And plug in the initial concentration and final concentration and volume to obtain:

V_1=\frac{2000mL*0.500M}{12.1M}\\\\V_1=82.6mL

Regards!

7 0
3 years ago
How many grams of NaOH would be required to make 1.0 L of a 1.5 M solution
omeli [17]
60 grams are required.

Hope this helped you!
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Preparation of ethanoic acid from 1,1,1-trichloroethane​
alekssr [168]
Anaerobic transformations of 1,1,1-trichloroethane (TCA), 1,1-dichloroethane (DCA), and chloroethane (CA) were studied with sludge from a lab-scale, municipal wastewater sludge digester. TCA was biologically transformed to DCA and CA and further to ethane by reductive dechlorination. TCA was also converted to acetic acid and 1,1-dichloroethene (11DCE) by cell-free extract. 11DCE was further biologically converted to ethene. This pathway was confirmed by transformation tests of TCA, DCA and CA, by tests with cell-free extract, and by chloride release during TCA degradation.
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2 years ago
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