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WARRIOR [948]
2 years ago
9

Starting with: Lithium atomic mass: 7 g/mol atomic number: 3 Use Lithium as a starting place for A, B, and C. What would happen

if: A. A proton was added to Lithium (7) B. An electron was removed from Lithium (7) C. A neutron was added to Lithium (7) D. A proton, an electron, and 2 neutrons were removed from Lithium(7). Include the symbol, element name, atomic number and atomic mass in your answer for each.
Chemistry
1 answer:
marshall27 [118]2 years ago
3 0

Answer:

a. _3^7Li

b. _{3}^{7}Li^+

c. _{3}^{8}Li

d. _{2}^{4}He^+

Explanation:

Given lithium with atomic mass (M = 7) and atomic number (Z = 3), we may write it in a standard notation placing the atomic mass as a superscript and the atomic number as a subscript:

_3^7Li

a. Atomic number corresponds to the total number of protons a species has. Adding a proton to lithium would increase the atomic number by 1 unit. Besides, changing the atomic number would also change the element, as each element has a unique atomic number.

It's also important to remember that mass of an atom consists of a sum of protons and neutrons, so we'd also have to add 1 to the mass number. Therefore, we would have some species X with:

_{3+1}^{7+1}X=_{4}^{8}X

Now we need to identify X finding an element with Z = 4 in the periodic table. This is beryllium:

_{4}^{8}Be

b. Standard lithium atom has an atomic number of 3, this means it has 3 protons. Since we have an atom, the positive charge provided by protons is equal to the negative charge provided by electrons. This implies that a neutral atom has the same number of protons and electrons.

We start with 3 electrons in lithium. Removing an electron would produce a lithium cation with a +1 charge, as we now have a net charge of +3 from protons and -2 from electrons.

Therefore, we may represent this as lithium cation with a +1 charge:

_{3}^{7}Li^+

c. Neutrons don't have a charge but they are a component of the total mass an atom has. Essentially, adding a neutron wouldn't change the overall charge (the atomic number) of lithium.

However, adding a neutron would cause the mass to increase by 1, as each neutron (as well as proton) weighs 1 atomic mass unit. Since the atomic number doesn't change, the element would still be lithium in that case:

_{3}^{7+1}Li=_{3}^{8}Li

d. Firstly, we remove a proton, this means the atomic number of Li decreases by 1 unit, as an atomic number is equal to the number of protons:

Z=3-1=2

The mass would also decrease by 1 unit, as mass contains the number of protons and neutrons:

M=7-1=6

That said, upon performing the first step, we have helium (atomic number of 2 in the periodic table) with a mass of 6:

_{2}^{6}He

Now, we also remove 2 neutrons. This means the mass decreases by 2 units and the charge of He remains constant, as no protons are involved in this step and neutrons don't change the charge of a nucleus:

_{2}^{6-2}He=_{2}^{4}He

Finally, an electron is removed. Removing an electron would produce a helium cation with charge +1, as we'd have 2 protons and 1 remaining electron upon removing 1 electron from helium that has a total of 2 electrons in its atomic state:

_{2}^{4}He^+

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By the stoichiometry of above reaction,

1 mole of _{40}^{91}\textrm{Zr} is produced by 1 mole _{41}^{91}\textrm{Nb}

So, 0.17 moles of _{40}^{91}\textrm{Zr} will be produced by = \frac{1}{1}\times 0.17=0.17\text{ moles of }_{40}^{91}\textrm{Nb}

Amount of _{82}^{212}\textrm{K}

decomposed will be = 0.17 moles

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a=\frac{a_o}{2^n}

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a = amount of reactant left after n-half lives = 0.024

a_o = Initial amount of the reactant = 0.194

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0.024=\frac{0.194}{2^n}

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