<span>Lyle stirs 5.0 grams of salt into a beaker of water. He then adds 15.0 grams of pure iodine to the mixture. If the total mass of the new mixture is 225 grams, what is the mass of the water
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Answer:
a)
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b)
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Explanation:
A)
Remember that positive number superscripts mean electrons lack and negative numbers mean electrons 'excess' (if we compare it with the neutral element). So, for the case of Fe2+ which is converted to Fe3+, we know that in Fe2+ there is a two electrons lack, while in Fe3+ there is a 3 electrons lack; it means that Fe2+ was converted to Fe3+ but releasing one electron:
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The same analysis is applied to Br2; Br2 is a molecule which is said to have a zero superscript because it is an apolar covalent bond; and it is converted to Br-, which, according to what I wrote above, means that there is a one electron excess. So, Br2 must have received an electron in order to change to Br-; but Br2 can't change to Br- as simple as that because Br2 is a molecule, not an atom; it is a molecule that has two Br atoms, so, Br2 must give two Br- ions as products, but receiving one electron for each one:
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b)
Applying the same, in Mg2+ there is a 2 electrons lack, and in Mg is not electron lack (its superscript is zero), so Mg must have released two electrons in order to change to Mg2+:
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Cr3+ has a 3 electrons lack, and Cr2+ a two electrons one, so, Cr3+ must receive an electron to convert to Cr2+:
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A chemical symbol represent the number of an element
Answer:
Average Atomic Mass = 10.81
Explanation:
Data Given:
Atomic Mass of ¹⁰X = 10 amu
Relative Intensity of ¹⁰X = 23 %
Atomic Mass of ¹¹X = 11 amu
Relative Intensity ¹¹X = 100 %
It means that if there are 123 atoms of this mystery element then then 23 will be of an isotope with atomic mass of 10 amu and the remaining 100 will be of an isotope with atomic mass of 11 amu.
So, we will now calculate the total mass of all elements as;
Total Mass = (23 × 10) + (100 × 11)
Total Mass = (230) + (1100)
Total Mass = 1330 amu
Then, we will calculate the average mass as,
Average Atomic Mass = 1330 / 123
Average Atomic Mass = 10.81
J. J Thompson’s atomic model