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3 years ago

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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In the following reaction, how many moles of CO2 will form if 10 moles of C3H4 are reacted? How many moles of O2 will also be co

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40 moles of O₂

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Explanation:

Given parameters:

Number of moles of C₃H₄ = 10moles

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Number of moles of CO₂ = ?

Solution:

The number of moles helps to understand and make quantitative measurements involving chemical reactions.

We start by solving this sort of problem by ensuring that the given equation is properly balanced;

C₃H₄ + 4O₂ → 3CO₂ + 2H₂O

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Now, we work from the known to unknown. We know the number of moles of C₃H₄ to be 10moles;

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Good chemistry question

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The molecular formula of the given compound is $\mathrm{H}_{2} \mathrm{CO}_{3}$ also known as Carbonic acid.

<h3>What is empirical formula and molecular formula?</h3>

The simplest whole-number ratio of the various atoms in a compound is represented by an empirical formula.

The precise number of various atom types present in a compound's molecule is indicated by the molecular formula.

Given that,

H = 3.3%

C = 19.3%

O = 77.4%

No. of moles of H = 3.3/1

No. of moles of H = 3.3

No. of moles of C = 19.3 / 12

No. of moles of C = 1.60

No. of moles of O = 77.4/16

No. of moles of O = 4.83

Therefore, the ratio of the atoms of C, H and O = 3.3 : 1.60 : 4.83

Divide by smallest value which you get =3.3 / 1.60 : 1.60 / 1.60 : 4.83 / 1.60

The ratio of the atoms of C, H and O = 2 : 1 : 3

So, the empirical formula is $\mathrm{H}_{2} \mathrm{CO}_{2}$

Let the molecular formula is $\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right) \mathrm{n}$

Then, molar mass $$=(2 \times 1+1 \times 12+3 \times 16) n\\$

Molar mass = 62n

As the question, 62 n = 60

n = 0.96 or n = 1 (rounded off to nearest ones)

So, the molecular formula is $\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right) 1=\mathrm{H}_{2} \mathrm{CO}_{3}$ i.e., the compound is Carbonic acid.

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Formula for sodium iodine​

Formula for sodium iodine