All categories

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

You might be interested in

State three acids, bases, and Salts commonly used in therapeutic processes

Delvig [45]

Acetic acid, phosphoric acid, hydrochloric acid.

sodium bicarbonate, magnesium hydroxide, aluminum hydroxide

calcium carbonate, sodium bicarbonate, magnesium sulfate.

3 0

3 years ago

Generally, when going down a group on the periodic table:

antiseptic1488 [7]

The correct answer is ionic radii increase.

The ionic radii decrease as one move across the periodic table, that is, from left to right, while the ionic radius increases as one move from top to bottom on the periodic table. As one moves down a group in the periodic table, the supplementary layers of electrons are being added that usually results in the increase of the ionic radius as one moves down the periodic table.

7 0

3 years ago

Match each organ or function with its body system. lungs heart pumps blood respiration provides oxygen to the blood arteries Res

anyanavicka [17]

Answer:

Explanation:

The respiratory system works with the circulatory system to provide this oxygen and to remove the waste products of metabolism. It also helps to regulate pH of the blood. Respiration is the sequence of events that results in the exchange of oxygen and carbon dioxide between the atmosphere and the body cells.

7 0

3 years ago

Please help me!! I’m confused and this teacher doesn’t know how to teach.

Marianna [84]

Answer:

chemical change bcoz it cannot be in the past form

8 0

3 years ago

In the first step of the Ostwald process for the synthesis of nitric acid, ammonia is converted to nitric oxide by the high-temp

Paul [167]

Solution: The given balanced equation is:

$4NH_3+5O_2\rightarrow 4NO+6H_2O$

If we have a hypothetical equation:

$A+2B\rightarrow 3C+5D$

Then the rate could be written as:

$rate=-\frac{\Delta [A]}{\Delta t}=-\frac{1}{2}\frac{\Delta [B]}{\Delta t}=\frac{1}{3}\frac{\Delta [C]}{\Delta t}=\frac{1}{5}\frac{\Delta [D]}{\Delta t}$

From above expression one thing could easily be noticed that the coefficients of all are inverted. Also, there is negative sign in front of reactants and positive sign in front of the products. Negative sign stands for rate of consumption where as positive sign stands for rate of formation.

Like the above example, we can write the rate for the given equation and it would be looking as:

$rate=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}=-\frac{1}{5}\frac{\Delta [O_2]}{\Delta t}=\frac{1}{4}\frac{\Delta [NO]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_2O]}{\Delta t}$

Now we can easily answer all the parts of the question.

(a) From above expression, the rate of consumption of $O_2$ related to rate of consumption of $NH_3$ as:

$-\frac{1}{5}\frac{\Delta [O_2]}{\Delta t}=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}$

multiply both sides by -5

$\frac{\Delta [O_2]}{\Delta t}=\frac{5}{4}\frac{\Delta [NH_3]}{\Delta t}$

So, rate of consumption of oxygen is $\frac{5}{4}$ the rate of consumption of ammonia.

(b) The relationship between rate of formation of NO to the rate of consumption of ammonia will be written as:

$\frac{1}{4}\frac{\Delta [NO]}{\Delta t}=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}$

Multiply both sides by 4

$\frac{\Delta [NO]}{\Delta t}=-\frac{\Delta [NH_3]}{\Delta t}$

So, rate of formation of NO equals to the rate of consumption of ammonia.

Now, the rate of formation of $H_2O$ to the rate of consumption of ammonia would be:

$\frac{1}{6}\frac{\Delta [H_2O]}{\Delta t}=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}$

Multiply both sides by 6

$\frac{\Delta [H_2O]}{\Delta t}=-\frac{6}{4}\frac{\Delta [NH_3]}{\Delta t}$

So, the rate of formation of $H_2O$ is $\frac{6}{4}$ times that is 1.5 times to the rate of consumption of ammonia.

5 0

3 years ago

Formula for sodium iodine​

Formula for sodium iodine