The diagram below represents a sodium atom bonding to a chlorine atom to form sodium chlorine.

Atoms form chemical bonds to make their outer electron shells more stable. ... An ionic bond, where one atom essentially donates an electron to another, forms when one atom becomes stable by losing its outer electrons and the other atoms become stable (usually by filling its valence shell) by gaining the electrons.

Explanation:

7 0

3 years ago

Calcium carbonate is often used as an antacid. Your stomach acid is composed of HCl at a pH of 1.5. If you ate toooo much Turkey

stiks02 [169]

<u>Answer:</u> 0.0237 g of calcium carbonate would be required to neutralize the given amount of HCl

<u>Explanation:</u>

pH is defined as the negative logarithm of hydrogen ion concentration present in the solution

$pH=-\log [H^+]$ .....(1)

Given value of pH = 1.5

Putting values in equation 1:

$1.5=-\log[H^+]$

$[H^+]=10^{(-1.5)}=0.0316M$

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

$\text{Molarity of solution}=\frac{\text{Number of moles of solute}\times 1000}{\text{Volume of solution (mL)}}$ .....(2)

We are given:

Volume of solution = 15.0 mL

Molarity of HCl = 0.0316 M

Putting values in equation 2:

$0.0316=\frac{\text{Moles of HCl}\times 1000}{15.0}\\\\\text{Moles of HCl}=\frac{0.0316\times 15.0}{1000}=4.74\times 10^{-4}mol$

The chemical equation for the reaction of HCl and calcium carbonate follows:

$2HCl+CaCO_3\rightarrow H_2CO_3+CaCl_2$

By the stoichiometry of the reaction:

2 moles of HCl reacts with 1 mole of calcium carbonate

So, $4.74\times 10^{-4}mol$ of HCl will react with = $\frac{1}{2}\times 4.74\times 10^{-4}=2.37\times 10^{-4}mol$ of calcium carbonate

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of calcium carbonate = $2.37\times 10^{-4}mol$

Molar mass of calcium carbonate = 100.01 g/mol

Putting values in the above equation:

$\text{Mass of }CaCO_3=(2.37\times 10^{-4}mol)\times 100.01g/mol\\\\\text{Mass of }CaCO_3=0.0237g$

Hence, 0.0237 g of calcium carbonate would be required to neutralize the given amount of HCl

7 0

2 years ago