A. new ocean ridges and seafloor
Answer:
6.564×10¹⁶ fg.
Explanation:
The following data were obtained from the question:
Mass of beaker = 76.9 g
Mass of beaker + salt = 142.54 g
Mass of salt in fg =?
Next, we shall determine the mass of the salt in grams (g). This can be obtained as follow:
Mass of beaker = 76.9 g
Mass of beaker + salt = 142.54 g
Mass of salt =?
Mass of salt = (Mass of beaker + salt) – (Mass of beaker)
Mass of salt = 142.54 – 76.9
Mass of salt = 65.64 g
Finally, we shall convert 65.64 g to femtograms (fg) as illustrated below:
Recall:
1 g = 1×10¹⁵ fg
Therefore,
65.64 g = 65.64 g × 1×10¹⁵ fg / 1g
65.64 g = 6.564×10¹⁶ fg
Therefore, the mass of the salt is 6.564×10¹⁶ fg.
<u>Answer:</u>
<u>For a:</u> The isotopic representation of iodine is 
<u>For b:</u> The isotopic representation of cesium is 
<u>For c:</u> The isotopic representation of strontium is 
<u>Explanation:</u>
The isotopic representation of an atom is: 
where,
Z = Atomic number of the atom
A = Mass number of the atom
X = Symbol of the atom
We are given:
Number of neutrons = 78
Atomic number of iodine = 53 = Number of protons
Mass number = 53 + 78 = 131
Thus, the isotopic representation of iodine is 
We are given:
Number of neutrons = 82
Atomic number of cesium = 55 = Number of protons
Mass number = 55 + 82 = 137
Thus, the isotopic representation of cesium is 
We are given:
Number of neutrons = 52
Atomic number of strontium = 38 = Number of protons
Mass number = 38 + 52 = 90
Thus, the isotopic representation of strontium is 
<span><span>LiF, LiCl, LiBr, LiI, LiAtNaF, NaCl, NaBr, NaI, NaAtKF, KCl, KBr, KI, KAt</span><span>RbF, RbCl, RbBr, RbI, RbAt CsF, CsCl, CsBr, CsI, CsAt FrF, FrCl, FrBr, FrI, FrAt<span>
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