Answer:
106.905 amu is the mass of the other isotope
Explanation:
The atomic mass of an element is the sum of the masses of the isotopes multiplied by its abundance. The atomic mass of an element X with 2 isotopes is:
X = X-109*i + X-107*i
Where X is the atomic mass = 107.868 amu
X-109 = 108.905amu, i = 48.16% = 0.4816
X-107 = ?, i = 1-0.4816 = 0.5184
Replacing:
107.868amu = 108.905amu*0.4816 + X-107*0.5184
55.4194 = X-107*0.5184
106.905 = X-107
<h3>106.905 amu is the mass of the other isotope</h3>
We find the weight of the empirical formula:
12.0107 + 2 x 1.00794 + 15.9994
= 30.03
Now, we divide the molecular weight by the weight of the empirical formula to find the number of times the empirical formula repeats:
90.09 / 30.03
= 3
The formula is 3(CH₂O)
C₃H₆O₃
Answer:
Plutonium is the second transuranium element of the actinide series.
Answer:
315.
Explanation:
Hello.
In this case, since the given number has five significant figures as the zero is to the right of the first nonzero digit (3), if it is required to report it with three significant figures, it is necessary to "cut" it at the first five without any rounding since the subsequent zero is less than five.
Thus the number turns out:
315
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