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Lana71 [14]
3 years ago
11

Fluorine has seven electrons in its outermost shell. it will most likely try to join an atom with

Chemistry
2 answers:
Elanso [62]3 years ago
6 0

<u>Answer</u><u>:</u>

F (Fluorine) has (8-1 = 7) electrons in outer most shell and so, it will try to bond with an atom which has only 1 electron in its outer most shell.

<u>Explanation:</u>  

According to Octet rule, most of the atoms are said to be stable if it has 8 electrons in its outermost shell. As already said, Fluorine atom has 7 electrons while it requires 1 more electron to attain stability.  

Hence, F atom will try to join another atom with only one electron in outer shell, like H, Li and Na.  

joja [24]3 years ago
5 0

Answer:

Its either with itself to become fluoride Fl2 or with an alkali metal (elements in group 1) to achieve a stable electronic configuration or octet structure

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Methane and chlorine react to form four products: CH3Cl, CH2Cl2, CHCl3, and CCl4. At a particular temperature and pressure, 38.4
Anastasy [175]

Answer:

How many grams of CCL4 were formed? 116.9 g

How many grams of Cl2 reacted with the CH4? 243.8 g

Explanation:

First we need to know the molar mass for every element or compound in the reaction:

M_{CH_{4}}=16 g/mol\\M_{CH_{3}Cl}=50.49g/mol\\M_{CH_{2}Cl_{2}}=84.93g/mol\\M_{CHCl_{3}}=119.38g/mol\\M_{CCl_{4}}=153.82g/mol

Now we proceed to calculate the amount of moles produced, per product:

n_{CH_{4}}=2.4\\n_{CH_{3}Cl}=0.18\\n_{CH_{2}Cl_{2}}=0.55\\n_{CHCl_{3}}=0.91\\n_{CCl_{4}}=n_{CH_{4}}-(n_{CH_{3}Cl}+n_{CH_{2}Cl_{2}}+n_{CHCl_{3}})\\n_{CCl_{4}}=0.76mol\\m_{CCl_{4}}=n_{CCl_{4}}*M_{CCl_{4}}\\m_{CCl_{4}}=116.9g

To calculate the mass of chlorine we just need to make a mass balance:

m_{CH_{4}}+m_{Cl_{2}}=m_{CH_{3}Cl}+m_{CH_{2}Cl_{2}}+m_{CHCl_{3}}+m_{CCl_{4}}\\m_{Cl_{2}}=m_{CH_{3}Cl}+m_{CH_{2}Cl_{2}}+m_{CHCl_{3}}+m_{CCl_{4}}-m_{CH_{4}}\\m_{Cl_{2}}=243.8g

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Determine what kind of diet this animal has by examining its teeth.
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That animal is a Carnivore.

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The skull is that of a Panther which is a carnivore. Hope this helps

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At equivalence there is no more HA and no more NaOH, for this particular reaction. So that means we have a beaker of NaA and H2O. The H2O contributes 1 x 10-7 M hydrogen ion and hydroxide ion. But NaA is completely soluble because group 1 ion compounds are always soluble. So NaA breaks apart in water and it just so happens to be in water. So now NaA is broken up. The Na+ doesn't change the pH but the A- does change the pH. Remember that the A anion is from a weak acid. That means it will easily attract a hydrogen ion if one is available. What do you know? The A anion is in a beaker of H+ ions! So the A- will attract H+ and become HA. When this happens, it leaves OH-, creating a basic solution, as shown below.

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