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3 years ago

Sodium(na) is a highly reactive metal and chlorine is a poisonous gas. Which compound do they form when they chemically combine?

Chemistry

2 answers:

Nastasia [14]3 years ago

7 0

It would form into NaCl because Na has a +1 charge while Cl has a -1 charge.

damaskus [11]3 years ago

3 0

Sodium lose one electron and convert into sodium ion Na+

Chlorine lose one electron and form Cl- ion

When these ions come together they from an ionic compound NaCl,

table salt

The ionic compound which consist ionic bond. The ionic bond is a type of bond which formed between positive and negative ions or metal, sodium and non-metal , chlorine

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An ____,or an element located in group 7A, will form a salt when reacted with a metal atom.

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Answer:

B. Halogen

Explanation:

Halogens are the family of chemical elements found in the group VIIA of the periodic table which means they possess seven (7) outer electrons. Some examples of halogens are Fluorine (F), Iodine (I), Bromine (Br), and Chlorine (Cl).

A halogen, or an element located in group 7A, will form a salt when reacted with a metal atom.

For example, when Chlorine reacts with Sodium it would form a salt known as Sodium Chloride.

Na + Cl ----> NaCl

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3 years ago

8. What volume (ml) of a 5.45 M lead nitrate solution must be diluted to 820.7 ml to make a

Ganezh [65]

212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

Answer:

Option A.

Explanation:

Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.

$C_{1} V_{1} =C_{2} V_{2}$

So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.

Then, $(5.45*V_{1} ) = (820.7*1.41)$

$V_{1}=\frac{820.7*1.41}{5.45}= 212.33 ml$

So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

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3 years ago