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Tems11 [23]
3 years ago
11

Formula for sodium iodine​

Chemistry
2 answers:
ivolga24 [154]3 years ago
7 0

Answer:

NaI

Explanation:

The formula for sodium iodide is NaI

It is a compound (ionic compound) formed by the chemical reaction between sodium, Na and iodine.

The sodium is a metal in group 1 with atomic number 11 and mass number 23 ,it has one valence electron which it gives out to form a cation of +1 charge and the iodine is in group 7(halogen family) with atomic number 57 and mass number 23 and it has 7 valence electrons which allows it attract 1 electron to form an anion(iodide) of charge -1.

The reaction of sodium ion with iodide ion form the sodium iodide where sodium releases its one valence electron and iodine receives it which makes it balance that is they react in ratio 1:1 and thus have a chemical formula of NaI.

NaI has atomic mass of 150g/mole that is 127 + 23 = 150

Where

Na=23g/mol, I = 127g/mol

NaI is used mainly as nutritional supplement

I hope this was helpful, Please mark as brainliest.

Vlad1618 [11]3 years ago
6 0
Nal is the formula for sodium iodine
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The increasing metallic character of metal is a measure of their ability to lose electrons.

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2 years ago
In their notebook you see that it takes 9 hours for a sixth of a 0.5M solution of BC2 to react. Unfortunately, you have somewher
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration of BC_2 that should used originally is C_Z_o = 0.4492M

Explanation:

     From the question we are told that

         The necessary elementary step is  

                  2BC_2 ----->4C + B_2

          The  time taken for sixth of 0.5 M of reactant to react t = 9 hr

           The time available is t_a = 3.5 hr

             The desired concentration to  remain C  = 0.42M

Let Z be the reactant ,   Y be the first product and X the second product

Generally the elementary rate  law is mathematically as

                    -r_Z = kC_Z^2 = - \frac{d C_Z}{dt}

Where k is the rate constant ,  C_Z is the concentration of Z

From the elementary rate law we see that the reaction is second order (This because the concentration of the reactant is raised to power 2 )

 For second  order reaction

            \frac{1}{C_Z}  - \frac{1}{C_Z_o}  = kt

Where C_Z_o is the initial concentration of Z which a value of   C_Z_o = 0.5M

       From the question we are told that it take  9 hours  for the concentration of  the reactant to become

                 C_Z =  C_Z_o - \frac{1}{6}  C_Z_o

                  C_Z = 0.5  - \frac{0.5}{6}

                       = 0.4167 M

So      

                     \frac{1}{0.4167}  - \frac{1}{0.50}  =  9 k

                          0.400 = 9 k

                =>    k = 0.044\  L/ mol \cdot hr^{-1}

  For   C_Z = 0.42M

                \frac{1}{0.42} - \frac{1}{C_Z_o}  = 3.5 * 0.044

                2.38 -  0.154  =    \frac{1}{C_Z_o}

                           2.226  =    \frac{1}{C_Z_o}

                            C_Z_o = \frac{1}{2.226}

                             C_Z_o = 0.4492M

                       

           

             

         

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marusya05 [52]

Answer:

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The elements listed are in the same group

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Answer:

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3 years ago
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