Answer:
a. neutral
b. salts
c. salt
Explanation:
Organic salts are a dense number of ionic compounds with innumerable characteristics. They are previously derived from an organic compound, which has undergone a transformation that allows it to be a carrier of a charge, and that in addition, its chemical identity depends on the associated ion.
Organic salts are usually stronger acids or bases than inorganic salts. This is because, for example, in the amine salts, it has a positive charge due to its bond with an additional hydrogen: A + -H. Then, in contact with a base, donate the proton to be a neutral compound again
RA + H + B => RA + HB
H belongs to A, but it is written as it is involved in the neutralization reaction.
On the other hand, RA + can be a large molecule, unable to form solids with a crystalline network stable enough with the hydroxyl anion or oxyhydrile OH–.
When this is so, salt RA + OH– behaves as a strong base; even as basic as NaOH or KOH
1.56 moles of N2 are needed to fill a 35 L tank at standard temperature and pressure. Details about moles can be found below.
<h3>How to calculate number of moles?</h3>
The number of moles of a substance can be calculated using the following formula:
PV = nRT
Where;
- P = pressure
- V = volume
- n = number of moles
- R = gas law constant
- T = temperature
At STP;
- T = 273K
- P = 1 atm
- R = 0.0821 Latm/molK
1 × 35 = n × 0.0821 × 273
35 = 22.41n
n = 35/22.41
n = 1.56mol
Therefore, 1.56 moles of N2 are needed to fill a 35 L tank at standard temperature and pressure.
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Answer:
A pregnant woman would consume 1.812 x 10⁻⁶ oz of mercury in a month if she ate the maximum recommended amount of fish.
Explanation:
0.302 oz mercury _____________ 1 x 10⁶ oz bluefish
x _____________ 6.00 oz bluefish
x = 1.812 x 10⁻⁶ oz mercury
A pregnant woman would consume 1.812 x 10⁻⁶ oz of mercury in a month if she ate the maximum recommended amount of fish.
Answer:
A) CH3CH2CH2CH2CH2CH2OH
Explanation:
For this question, we have the following answer options:
A) CH3CH2CH2CH2CH2CH2OH
B) (CH3CH2)2CH(OH)CH2CH3
C) (CH3CH2)2CHOHCH3
D) (CH3CH2)3COH
E) (CH3CH2)2C(CH3)OH
We have to remember the<u> reaction mechanism</u> of the substitution reaction with 
. <em>The idea is to generate a better leaving group in order to add a "Br" atom.</em>
The attacks the "OH" generation new a bond to P (O-P bonds are very strong), due to this new bond we will have a better leaving group that can remove the oxygen an allow the attack of the Br atom to generating a new C-Br bond. This is made by an <u>Sn2 reaction</u>. Therefore we will have a faster reaction with <u>primary substrates</u>. In this case, the only primary substrate is molecule A. So, <em>"CH3CH2CH2CH2CH2CH2OH"</em> will react faster.
See figure 1
I hope it helps!
<span>First - you need the empirical formula.
So, assume you have 100 g of the compound.
If so, you'll have 54.53 gram of C, 9.15 g of H and 36.32 g of O. Find the number of moles of each.
54.53 g C (1 mole C / 12.01 g C) = 4.540
9.15 g H (1 mole H / 1.008 g H) = 9.077
36.32 g O (1 mole O / 15.9994 g O) = 2.270
Take the smallest number found and divide the others by it to get the empirical formula.
4.540/2.270 = 2.
9.077/2.270 = 4.
2.270/2.270 =1.
So, that gives you the empirical formula of C2H4O.
Find the weight of this compound. C = 12, H = 1, O = 16. So, C2H4O is 44 amu.
132/44 = 3.
So, 3 (C2 H4 O) = C6H12O3 = molecular formula.</span>
