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sergeinik [125]
3 years ago
5

Select the correct answer. When an atom in a reactant loses electrons, what happens to its oxidation number? A. Its oxidation nu

mber decreases. B. Its oxidation number doubles. C. Its oxidation number increases. D. Its oxidation number increases by one. E. Its oxidation number stays the same.
Chemistry
2 answers:
vfiekz [6]3 years ago
8 0

Answer:

C. Its oxidation number increases.

Explanation:

  • <em><u>Oxidation is defined as the loss of electrons by an atom while reduction is the gain of electrons by an atom</u></em>.
  • Atoms of elements have an oxidation number of Zero in their elemental state.
  • When an atom looses electrons it undergoes oxidation and its oxidation number increases.
  • For example, <em><u>an atom of sodium (Na) at its elemental state has an oxidation number of 0. When the sodium atom looses an electrons it becomes a cation, Na+, with an oxidation number of +1 , the loss of electron shows an increase in oxidation number from 0 to +1.</u></em>
xxTIMURxx [149]3 years ago
5 0

Answer:

When oxidation occurs, an element loses electrons and its oxidation number increases (becomes more positive). When reduction occurs, an element gains electrons and its oxidation number decreases or is reduced (becomes more negative).

Explanation:

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Answer:

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Explanation:

6 0
3 years ago
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I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r
maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P

Now, the moles of Li3P consumed by 15 g of Al2O3:

n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP

Therefore, the total mass of products is:

m_{products}=13g+8.5g\\\\m_{products}=21.5g

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

7 0
3 years ago
Why don’t we include the mass of an atoms electrons in the atomic mass?
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Because Electrons have a negative charge

8 0
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Answer:

30.83 M

Explanation:

17.03052 re in one mole. So, if you multiply it by 30.83, you will get 535 g of ammonia.

In fact, the detailed answer is 30.827009392549122.

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