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How many hydrogen ions are in 1 liter of stomach acid? (Hint: Avogadro’s number). Show your work and remember to put your answer

ch4aika [34]

Answer:

96,532,000,000,000,000,000,000 ions.

Explanation:

Assume the highest concentration of HCl in the stomach (160 mmol/L) that is in the canaliculi.

This means that 1.0 L of the stomach acid contains 0.160 mol.

Every 1.0 mole of HCl contains Avogadro’s number of molecules (6.022 x 10²³) molecules.

Every 1.0 molecule of HCl contains 1.0 H⁺ ion.

Using cross multiplication:

1.0 mole of HCl → 6.022 x 10²³ H⁺ ions.

0.160 mole of HCl → ??? H⁺ ions.

∴ The no. of H⁺ ions in 1 liter of stomach acid = (6.022 x 10²³ H⁺ ions)(0.160 mole of HCl) / (1.0 mole of HCl) = 9.6352 x 10²² ions = 96,532,000,000,000,000,000,000 ions.

3 0

3 years ago

Tire manufacturers design tires to stop quickly in various conditions. Use the table to answer the question: Which tire would be

Llana [10]

Can you show the picture?

7 0

3 years ago

Read 2 more answers

Given the reaction: Mg(s) + 2 AgNO3(aq) → Mg(NO3)2(aq) + 2 Ag(s) Which type of reaction is represented?

Papessa [141]

Mg(s) + 2 AgNO₃(aq) = Mg(NO₃)₂(aq) + 2 Ag(s)

reaction is : single replacement

hope this helps!

6 0

4 years ago

What is the pH of a buffer solution upon mixing 15.0 mL of 0.40 M HCl and 20.0 mL of 0.50 M NH? Kb (NH3) = 1.8 x 10 E. 7.00 A. 9

vladimir2022 [97]

Answer: The pH of resulting solution is 9.08

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ ........(1)

For HCl:

Molarity of HCl = 0.40 M

Volume of solution = 15.0 mL

Putting values in equation 1, we get:

$0.40M=\frac{\text{Moles of HCl}\times 1000}{15.0mL}\\\\\text{Moles of HCl}=0.006mol$

For ammonia:

Molarity of ammonia = 0.50 M

Volume of solution = 20.0 mL

Putting values in equation 1, we get:

$0.50M=\frac{\text{Moles of ammonia}\times 1000}{20.0mL}\\\\\text{Moles of ammonia}=0.01mol$

The chemical reaction for hydrochloric acid and ammonia follows the equation:

$HCl+NH_3\rightarrow NH_4Cl$

Initial: 0.006 0.01

Final: - 0.004 0.006

Volume of solution = 15.0 + 20.0 = 35.0 mL = 0.035 L (Conversion factor: 1 L = 1000 mL)

To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:

$pOH=pK_b+\log(\frac{[salt]}{[base]})$

$pOH=pK_b+\log(\frac{[NH_4Cl]}{[NH_3]})$

We are given:

$pK_b$ = negative logarithm of base dissociation constant of ammonia = $-\log (1.8\times 10^{-5})=4.74$

$[NH_4Cl]=\frac{0.006}{0.035}$

$[NH_3]=\frac{0.004}{0.035}$

pOH = ?

Putting values in above equation, we get:

$pOH=4.74+\log(\frac{0.006/0.035}{0.004/0.035})\\\\pOH=4.92$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH=14-4.92=9.08$

Hence, the pH of the solution is 9.08

3 0

4 years ago

Plz!!!!!!!!!!!!!!!!!help!!!!!!!!!!!!!!!!!!!!

drek231 [11]

An electromagent is being shown. Because the picture is showing how to put the lightnulb in and where it is

8 0

3 years ago