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Andrews [41]
3 years ago
7

Based on the information in the table, which elements are most likely in the same periods of the periodic table?

Chemistry
2 answers:
daser333 [38]3 years ago
3 0
The answer is A.<span>Boron and carbon are likely together in one period because they have very close atomic numbers, while gallium and germanium are likely together in another period because they have very close atomic numbers.
</span>
Stells [14]3 years ago
3 0

The element that  are most likely in the same period of periodic table is

<em> Boron  and carbon  are likely together in one period  because they  have  very close  atomic   number, while  gallium and germanium  are likely  together  in another period  because  they  have  very  close atomic  numbers ( answer A)</em>


<u>Explanation</u>

  • Element  in   same period  have the  close atomic  number .
  • The   atomic number of  an element is determined by number of protons the element have.
  • The number of proton are equal to number of electrons

Therefore   boron is in atomic number  5  with  2.3  electronic  configuration  while  carbon  is in atomic number 6 with 2.4  electron  configuration

since both  boron  and carbon  have two  energy level  they are in period 2.


Gallium  is in atomic number 31 with  2.8.8.8.5  electron configuration while Germanium (Ge)  is in atomic number 32 with  2.8.8.8.6  electron configuration.

Since  both gallium  and germanium  has 5  energy level they are in  same period (period 5)


  • for example  Aluminum,boron  and gallium are in group (iii)  since they have  3 valence electrons.
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calculate the mass of calcium phosphate and the mass of sodium chloride that could be formed when a solution containing 12.00g o
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Answer : The mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

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First we have to calculate the moles of Na_3PO_4 and CaCl_2.

\text{Moles of }Na_3PO_4=\frac{\text{Given mass }Na_3PO_4}{\text{Molar mass }Na_3PO_4}

\text{Moles of }Na_3PO_4=\frac{12.00g}{164g/mol}=0.0732mol

and,

\text{Moles of }CaCl_2=\frac{\text{Given mass }CaCl_2}{\text{Molar mass }CaCl_2}

\text{Moles of }CaCl_2=\frac{10.0g}{111g/mol}=0.0901mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is:

2Na_3PO_4+3CaCl_2\rightarrow 6NaCl+Ca_3(PO_4)_2

From the balanced reaction we conclude that

As, 3 mole of CaCl_2 react with 2 mole of Na_3PO_4

So, 0.0901 moles of CaCl_2 react with \frac{2}{3}\times 0.0901=0.0601 moles of Na_3PO_4

From this we conclude that, Na_3PO_4 is an excess reagent because the given moles are greater than the required moles and CaCl_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NaCl  and Ca_3(PO_4)_2

From the reaction, we conclude that

As, 3 mole of CaCl_2 react to give 6 mole of NaCl

So, 0.0901 mole of CaCl_2 react to give \frac{6}{3}\times 0.0901=0.1802 mole of NaCl

and,

As, 3 mole of CaCl_2 react to give 1 mole of Ca_3(PO_4)_2

So, 0.0901 mole of CaCl_2 react to give \frac{1}{3}\times 0.0901=0.030 mole of Ca_3(PO_4)_2

Now we have to calculate the mass of NaCl  and Ca_3(PO_4)_2

\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl

\text{ Mass of }NaCl=(0.1802moles)\times (58.5g/mole)=10.5g

and,

\text{ Mass of }Ca_3(PO_4)_2=\text{ Moles of }Ca_3(PO_4)_2\times \text{ Molar mass of }Ca_3(PO_4)_2

\text{ Mass of }Ca_3(PO_4)_2=(0.030moles)\times (310g/mole)=9.3g

Therefore, the mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

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