Oxidation state number can be used to determine the unknown element in these two compounds. They are used to determine how many electrons are given, taken or shared to form compounds.
Recall the elementary rules of oxidation numbers.
1. The sum of all oxidation numbers in a neutral compound is zero.
2. Chlorine, bromine, iodine have oxidation number of -1 ( except compounds with fluorine and oxygen)
Let oxidation number of element M be x.
Check rule 2. Chlorine has -1 oxidation number.
Now we write an equation of MCl₂ (neutral compound)
x + (2 * -1)= 0 ⇒ x₁= +2
For MCl₃
x + (3 * -1)= 0 ⇒ x₂= +3
So element has 2 different oxidation number in compounds, +2 and +3.
The element is iron (Fe) since it has +2 and +3 oxidation numbers in the compounds.
You need to learn it by hard. Unfortunately there is not an easier way to work out with these oxidation numbers.
The answer is iron (Fe).
(a) In this section, give your answers to three decimal places.
(i)
Calculate the mass of carbon present in 0.352 g of CO
2
.
Use this value to calculate the amount, in moles, of carbon atoms present in 0.240 g
of
A
.
(ii)
Calculate the mass of hydrogen present in 0.144 g of H
2
O.
Use this value to calculate the amount, in moles, of hydrogen atoms present in 0.240 g
of
A
.
(iii)
Use your answers to calculate the mass of oxygen present in 0.240 g of
A
Use this value to calculate the amount, in moles, of oxygen atoms present in 0.240 g
of
A
(b)
Use your answers to
(a)
to calculate the empirical formula of
A
thank you
hope it helpsss
Answer:
(A) pH < 1 the predominant form is the cation: H3C-C(H)(NH3+)-COOH
(B) pH = pl the predominant form is the zwitterion H3C-C(H)(NH3+)-COO-
(C) pH > 11 the predominant form is the anion: H3C-C(H)(NH2)-COO-
(D) Does not occurs in any significant pH: H3C-C(H)(NH2)-COOH
Explanation:
Amino acids are bifunctional because they have an amine group and a carboxyl group. The amine group is a weak base and the carboxyl group is a weak acid, but the pKa of both groups will depend on the whole structure of the amino acid. Also, every amino acid has an isoelectric point (pI), which means the pH were the predominant form of the amino acid is the zwitterion. The structure of the alanine (CH3CH2NH2COOH) shows it has the carboxyl group at C1 with a pKa1 of 2.3 and the amino group at C2 whit the pKa2 of 9.7. The isoelectric poin (pI) of Alanine is 6. Consequently, the protonation of the molecule will depend on the pH of the solution. There are three possibilities:
1) If the pH is under the pKa of the carboxyl group (2.3) the predominant form will be with the amino group protonated, forming a cation (CH3CH(NH3+)COOH).
2) If the pH is between pKa1 (2.3) and pKa2 (9.7) the predominant form will be the zwitterion (CH3CH(NH3+)(COO-)).
3) If the pH is upper the pKa2 of the amino group (9.7) the predominant form will be with the carboxyl group deprotonated, forming an anion (CH3CHNH2(COO-)).
Answer:
The partial pressure of neon in the vessel was 239 torr.
Explanation:
In all cases involving gas mixtures, the total gas pressure is related to the partial pressures, that is, the pressures of the individual gaseous components of the mixture. Put simply, the partial pressure of a gas is the pressure it exerts on a mixture of gases.
Dalton's law states that the total pressure of a mixture of gases is equal to the sum of the pressures that each gas would exert if it were alone. Then:
PT= P1 + P2 + P3 + P4…+ Pn
where n is the amount of gases present in the mixture.
In this case:
PT=PN₂ + PAr + PHe + PNe
where:
- PT= 987 torr
- PN₂= 44 torr
- PAr= 486 torr
- PHe= 218 torr
- PNe= ?
Replacing:
987 torr= 44 torr + 486 torr + 218 torr + PNe
Solving:
987 torr= 748 torr + PNe
PNe= 987 torr - 748 torr
PNe= 239 torr
<u><em>The partial pressure of neon in the vessel was 239 torr.</em></u>
Answer:
1219.5 kj/mol
Explanation:
To reach this result, you must use the formula:
ΔHºrxn = Σn * (BE reactant) - Σn * (BE product)
ΔHºrxn = [1 * (BE C = C) + 2 * (BE C-H) + 5/2 * (BE O = O)] - [4 * (BE C = O) + 2 * (BE O-H).
The BE values are:
BE C = C: 839 kj / mol
BE C-H: 413 Kj / mol
BE O = O: 495 kj / mol
BE C = O = 799 Kj / mol
BE O-H = 463 kj / mol
Now you must replace the values in the above equation, the result of which will be:
ΔHºrxn = [1 * 839 + 2 * (413) + 5/2 * (495)] - [4 * (799) + 2 * (463) = 1219.5 kj/mol
