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3 years ago

Inorganic compounds which are found in the earth are called: minerals vitamins synthetics cement

Chemistry

2 answers:

Dominik [7]3 years ago

6 0

Answer:

Minerals

Explanation:

I looked it up on google. Thanks.

vampirchik [111]3 years ago

5 0

Answer:

synthetic cement

Explanation: Not found naturally in nature, unlike minerals and vitamins

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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

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Charra [1.4K]

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