<u>Answer:</u> The lewis dot structure is attached below.
<u>Explanation:</u>
A Lewis dot structure is defined as the representation of atoms having electrons around the atom where electrons are represented as dots.
A ketene is an organic compound having general formula R′R″C=C=O, where R and R' are two different/same monovalent chemical groups.
The given chemical compound having formula 
is represented as 
.
Total number of unshared electrons = 4 (left on oxygen atom only)
The lewis dot structure of is given in the image below.
First write all of the compounds/atoms in either side then fill in existing values and balance
Na- 1
Br- 1
Ca- 1
Cl- 2
Na- 1
Cl- 1
Ca-1
Br-2
Balance to get
2NaBr+CaCl2=2NaCl+CaBr2
Hello! I can help you with this. First, convert them into it’s written out standard form. 10^4 is 10,000. 10,00 * 1.26 is 12,600. 10,000 * 2.5 is 25,000. 12,600 + 25,000 = 37,600 or 3.76 * 10^4 in scientific notation. The answer in scientific notation is 3.76 * 10^4.
Answer: chemical
not 100% sure
Answer:
Mostly Para
Explanation:
First, let's assume that the molecule is the toluene (A benzene with a methyl group as radical).
Now the nitration reaction is a reaction in which the nitric acid in presence of sulfuric acid, react with the benzene molecule, to introduce the nitro group into the molecule. The nitro group is a relative strong deactiviting group and is metha director, so, further reactions that occur will be in the metha position.
Now, in this case, the methyl group is a weak activating group in the molecule of benzene, and is always ortho and para director for the simple fact that this molecule (The methyl group) is a donor of electrons instead of atracting group of electrons. Therefore for these two reasons, when the nitration occurs,it will go to the ortho or para position.
Now which position will prefer to go? it's true it can go either ortho or para, however, let's use the steric hindrance principle. Although the methyl group it's not a very voluminous and big molecule, it still exerts a little steric hindrance, and the nitro group would rather go to a position where no molecule is present so it can attach easily. It's like you have two doors that lead to the same place, but in one door you have a kid in the middle and the other door is free to go, you'll rather pass by the door which is free instead of the door with the kid in the middle even though you can pass for that door too. Same thing happens here. Therefore the correct option will be mostly para.
