The first answer choice is correct: SF6 is a compound: it is a substance composed of more than one atom from different elements (sulfur and fluorine).
The second answer choice is incorrect: while SF6 is indeed a compound that contains seven atoms, those atoms are not identical since one is a sulfur atom and six are fluoride atoms.
The third answer choice is incorrect: SF6 is not an element because it can be separated chemically into simpler substances that are elements. All the atoms comprising an element must have the same number of protons (the same atomic number); sulfur and fluorine have different atomic numbers. There is no single nucleus in SF6; the S and six F atoms each has their own nucleus.
The fourth answer choice is incorrect; SF6 contains no carbon atom, and so wouldn’t meet a central criterion for an organic compound. Moreover, the formula SF6 indicates that the sulfur atom is not bonded to any other atom apart from fluorine atoms, and there are no bonding electrons left on the S. There is no plausible reason to think that a molecule with such weak intermolecular attractions as SF6 would form long chains of S atoms. Indeed, in standard conditions, SF6 is a gas.
Water
It can be a solid or a liquid
L'air n'est pas un mélange gazeux, donc la réponse serait le verre
Answer: The molarity of each of the given solutions is:
(a) 1.38 M
(b) 0.94 M
(c) 1.182 M
Explanation:
Molarity is the number of moles of a substance present in liter of a solution.
And, moles is the mass of a substance divided by its molar mass.
(a) Moles of ethanol (molar mass = 46 g/mol) is as follows.
Now, molarity of ethanol solution is as follows.
(b) Moles of sucrose (molar mass = 342.3 g/mol) is as follows.
Now, molarity of sucrose solution is as follows.
(c) Moles of sodium chloride (molar mass = 58.44 g/mol) are as follows.
Now, molarity of sodium chloride solution is as follows.
Thus, we can conclude that the molarity of each of the given solutions is:
(a) 1.38 M
(b) 0.94 M
(c) 1.182 M
Answer:
5. +5
6. +7
Explanation:
The oxidation state can be obtained as follow:
5. Oxidation state of oxygen, O = –2
Oxidation state of phosphorus, P =?
P3O10 = –5
3P + 10O = –5
3P + (10 x –2) = –5
3P – 20 = –5
Collect like terms
3P = –5 + 20
3P = 15
Divide both side by 3
P = 15/3
P = +5
Therefore, the oxidation state of P in P3O10^5- is +5
6. Oxidation state of oxygen, O = –2
Oxidation state of manganese, Mn =.?
Mn2O7 = 0
2Mn + 7O = 0
2Mn + (7 x –2) = 0
2Mn – 14 = 0
Collect like terms
2Mn = 14
Divide both side by 2
Mn = 14/2
Mn = +7
Therefore, the oxidation state of Mn in Mn2O7 is +7
