Chemical composition and function of iodized salt

Chemistry

1 answer:

Nutka1998 [239]2 years ago

7 0

Answer:

What is the chemical composition of iodized salt?

Iodized salt is a very cheap and good way of getting the necessary iodine. Ordinary table (cooking) salt is a chemical compound called "sodium chloride" (its chemical formula is NaCl). "Sodium iodide" (NaI) is added to the NaCl to make iodized salt.

Salt that has been iodized contains trace levels of sodium or potassium iodide. It's just regular salt sprinkled with potassium iodate. It has the same appearance and flavor! The bulk of table salt used currently is iodized, which has a number of advantages.

function of iodized salt
Iodized salt aids in the production of hormones that control heart rate and blood pressure. It also aids in the burning of unwanted fat deposits that may lead to heart disease. Salt maintains a healthy amount of hydration and electrolyte balance.

You might be interested in

Given the balanced chemical equation, SiO2(s) + 4 HF(g) → SiF4(g) + 2 H2O(l) ΔH°rxn = −184 kJ Determine the mass (in grams) of H

Ostrovityanka [42]

Answer:

mass HF = 150.05 g

Explanation:

SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l)

⇒ Q = (ΔH°rxn * mHF) / (mol HF * MwHF )

∴ MwHF = 20.0063 g/mol

∴ mol HF = 4 mol

∴ ΔH°rxn = - 184 KJ

∴ Q = 345 KJ

mass HF ( mHF ):

⇒ mHF = ( Q * mol HF * MwHF ) / ΔH°rxn

⇒ mHF = ( 345 KJ * 4mol HF * 20.0063 g/mol ) / 184 KJ

⇒ mHF = 150.05 g HF

3 0

3 years ago

1. Complete the following

Marat540 [252]

Answer:

$\large \boxed{\text{0.603 mol}}$

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of O₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 180.16

C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g: 18.1

(a) Moles of C₆H₁₂O₆

$\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{18.1 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1005 mol C$_{6}$H$_{12}$O}_{6}$

b) Moles of O₂

$\text{Moles of O}_{2} =\text{0.1005 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol O}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.603 mol O}_{2}\\\\\text{The reaction requires $\large \boxed{\textbf{0.603 mol}}$ of oxygen}$