Answer:
a mixture of two these
Explanation:
The number of isomeric monochlorides depends on the structure and number of equivalent hydrogen atoms in each isomer of pentane.
n-pentane has three different kinds of equivalent hydrogen atoms leading to three isomeric monochlorides formed.
Isopentane has four different types of equivalent hydrogen atoms hence four isomeric monochlorides are formed.
Lastly, neopentane has only one type of equivalent hydrogen atoms that yields one mono chlorination product.
Hence the cylinder must contain a mixture of isopentane and neopentane which yields four and one isomeric monochlorides giving a total of five identifiable monochloride products as stated in the question.
Answer:
The rate of disappearance of C₂H₆O = 2.46 mol/min
Explanation:
The equation of the reaction is given below:
2 K₂Cr₂O₇ + 8 H₂SO₄ + 3 C₂H₆O → 2 Cr₂(SO₄)₃ + 2 K₂SO₄ + 11 H₂O
From the equation of the reaction, 3 moles of C₂H₆O is used when 2 moles of Cr₂(SO₄)₃ are produced, therefore, the mole ratio of C₂H₆O to Cr₂(SO₄)₃ is 3:2.
The rate of appearance of Cr₂(SO₄)₃ in that particular moment is given 1.64 mol/min. This would than means that C₂H₆O must be used up at a rate which is approximately equal to their mole ratios. Thus, the rate of of the disappearance of C₂H₆O can be calculated from the mole ratio of Cr₂(SO₄)₃ and C₂H₆O.
Rate of disappearance of C₂H₆O = 1.64 mol/min of Cr₂(SO₄)₃ * 3 moles of C₂H₆O / 2 moles of Cr₂(SO₄)₃
Rate of disappearance of C₂H₆O = 2.46 mol/min of C₂H₆O
Therefore, the rate of disappearance of C₂H₆O = 2.46 mol/min
Number of moles = mass of Ni /molecular mass of Ni
mass of nickel = 86.4 g
molecular mass of nickel = 58.69
number of moles of Ni in 86.4 g
=86.4/58.69
=1.472 mol
(rounded to four significant figures instead of three because the first digit of the answer starts with a 1).
44. (a) N2O3 (b) SF4 (c) AlCl3 (d) Li2CO3
46. H Br
δ+ δ−
48. The metallic potassium atoms lose one electron and form +1 cations,
and the nonmetallic fluorine atoms gain one electron and form –1 anions.
K → K+
+ e–
19p/19e–
19p/18e–
F + e–
→ F–
9p/9e–
9p/10e–
The ionic bonds are the attractions between K+
cations and F–
anions.
50. See Figure 3.6.
52. (a) covalent…nonmetal-nonmetal (b) ionic…metal-nonmetal
54. (a) all nonmetallic atoms - molecular (b) metal-nonmetal - ionic
56. (a) 7 (b) 4
58. Each of the following answers is based on the assumption that nonmetallic
atoms tend to form covalent bonds in order to get an octet (8) of
electrons around each atom, like the very stable noble gases (other than
helium). Covalent bonds (represented by lines in Lewis structures) and lone
pairs each contribute two electrons to the octet.
(a) oxygen, O
If oxygen atoms form two covalent bonds, they will have an octet of electrons
around them. Water is an example:
H O H
(b) fluorine, F
If fluorine atoms form one covalent bond, they will have an octet of electrons
around them. Hydrogen fluoride, HF, is an example:
H F
(c) carbon, C
If carbon atoms form four covalent bonds, they will have an octet of electrons
around them. Methane, CH4, is an example:
H H
H
H
C
(d) phosphorus, P
If phosphorus atoms form three covalent bonds, they will have an octet
