51.86 grams would be in the container.
One ounce is an equivalent of 28.34 grams, so times that by 12.2
.
12.2 * 28.34 = 345.75.
Put the percentage into decimal form, so 15% would now be 0.15.
0.15 * 345.75 = 51.86.
Answer:
3
Explanation:
the answer is 3 because it is 3 for the o2 so 3 you <em>have </em><em>to </em><em>pay </em><em>more </em><em>attention </em><em>for </em><em>the </em><em>small </em><em>ditails </em>
Answer:
(a) sp³ sp³
H₃<u>C</u> - <u>C</u>H₃
(b) sp³ sp²
H₃<u>C</u> - <u>C</u>H = <u>C</u>H₂
sp²
(c) sp³ sp
H₃<u>C</u> - <u>C</u> ≡ <u>C</u> - <u>C</u>H₂OH
sp sp³
(d) sp³ sp²
H₃<u>C</u> - <u>C</u>H=O
Explanation:
Alkanes or the carbons with all the single bonds are sp³ hybridized.
Alkenes or the carbons with double bond(s) are sp² hybridized.
Alkynes or the carbons with triple bond are sp hybridized.
Considering:
(a) H₃C-CH₃ , Both the carbons are bonded by single bond so both the carbons are sp³ hybridized.
Hence,
sp³ sp³
H₃<u>C</u> - <u>C</u>H₃
(b) H₃C-CH=CH₂ , The carbon of the methyl group is sp³ hybridized as it is boned via single bonds. The rest 2 carbons are sp² hybridized because they are bonded by double bond.
Hence,
sp³ sp²
H₃<u>C</u> - <u>C</u>H = <u>C</u>H₂
sp²
(c) H₃C-C≡C-CH₂OH , The carbons of the methyl group and alcoholic group are sp³ hybridized as it is boned via single bonds. The rest 2 carbons are sp hybridized because they are bonded by triple bond.
Hence,
sp³ sp
H₃<u>C</u> - <u>C</u> ≡ <u>C</u> - <u>C</u>H₂OH
sp sp³
(d)CH₃CH=O, The carbon of the methyl group is sp³ hybridized as it is boned via single bonds. The other carbon is sp² hybridized because it is bonded by double bond to oxygen.
Hence,
sp³ sp²
H₃<u>C</u> - <u>C</u>H=O
Answer:
0
Explanation:
There are no unpaired electrons in the given element. It must be noted that for the atom above, we have even numbered electrons. The total electron we are having here is 18.
Now, we must also know that while the s orbital is not degenerate, the P orbital is degenerate. What this mean is that the p orbital is broken down into three different sub orbitals which is the Px , Py and Pz. Hence we can see that there are 6 electrons to enter into the P orbital too.
We can see that all the S orbitals have been completely filled with two electrons alike each. This is also the case for the P orbital as the 3 suborbitals take in 2 each to give a total of six
E. Double Replacement
The OH replaces the Cl, and the Cl replaces the OH
