Electrophoresis is a genetic engineering technique that can be used to separate segments of

The daily output of stomach acid (gastric juice) is 1000 to 2000 mL. Prior to a meal, stomach acid (HCl) typically has a pH of 1

Semmy [17]

Answer:

B.) 2 HCl(aq) + CaCO₃(s) → CaCl₂(aq) + H₂O(l) + CO₂(g)

C.) 7.4 × 10² mL

D.) 2 HCl(aq) + Mg(OH)₂(aq) → MgCl₂(aq) + H₂O(l)

E.) 1.3 × 10³ mL

Explanation:

The daily output of stomach acid (gastric juice) is 1000 to 2000 mL. Prior to a meal, stomach acid (HCl) typically has a pH of 1.49.

We can calculate the concentration of H⁺ using the definition of pH.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog (-1.49) = 0.0324 M

Since HCl is a strong monoprotic acid, we can consider this to be the concentration of HCl as well.

B.) One chewable tablet of the antacid Maalox contains 600 mg of CaCO₃. Enter the neutralization equation. Express your answer as a molecular equation including phases.

The neutralization equation is:

2 HCl(aq) + CaCO₃(s) → CaCl₂(aq) + H₂O(l) + CO₂(g)

C.) Given that one chewable tablet of the antacid Maalox contains 600 mg of CaCO₃, calculate the milliliters of stomach acid neutralized by two tablets of Maalox. Express the volume in milliliters to two significant figures.

We can establish the following relations:

Each tablet has 600 mg (0.600 g) of CaCO₃.
The molar mass of CaCO₃ is 100.09 g/mol.
The molar ratio of HCl to CaCO₃ is 2:1.
The concentration of HCl is 0.0324 mol/L.

The mililiters of HCl that neutralize 2 tablets of Maalox are:

$2Tablet.\frac{0.600gCaCO_{3}}{1Tablet} .\frac{1molCaCO_{3}}{100.09gCaCO_{3}} .\frac{2molHCl}{1molCaCO_{3}} .\frac{1000mLHCl}{0.0324molHCl} =7.4 \times 10^{2} mLHCl$

D.) The antacid milk of magnesia contains 400 mg of Mg(OH)₂ per teaspoon. Enter the neutralization equation. Express your answer as a molecular equation including phases.

The neutralization equation is:

2 HCl(aq) + Mg(OH)₂(aq) → MgCl₂(aq) + H₂O(l)

E.) Given that the antacid milk of magnesia contains 400 mg of Mg(OH)₂ per teaspoon, calculate the number of milliliters of stomach acid that are neutralized by 1 tablespoon of milk of magnesia. (1 tablespoon = 3 teaspoons.)Express the volume in milliliters to two significant figures.

We can establish the following relations:

1 tablespoon = 3 teaspoons
1 teaspoon contains 400 mg (0.400 g) of Mg(OH)₂
The molar mass of Mg(OH)₂ is 58.32 g/mol.
The molar ratio of HCl to Mg(OH)₂ is 2:1.
The concentration of HCl is 0.0324 mol/L.

The mililiters of HCl that neutralize 1 tablespoon of milf of magnesia are:

$1Tablespoon.\frac{3Teaspoon}{1Tablespoon} .\frac{0.400gMg(OH)_{2}}{1Teaspoon} .\frac{1molMg(OH)_{2}}{58.32gMg(OH)_{2}} .\frac{2molHCl}{1molMg(OH)_{2}} .\frac{1000mLHCl}{0.0324molHCl} =1.3 \times 10^{3} mLHCl$

4 0

3 years ago

This solution contains equal concentrations of both hf(aq) and naf(aq). write the proton transfer equilibrium (net ionic) reacti

Viktor [21]

Net ionic reaction for the solution containing equal concentrations of both HF(aq) and NaF(aq) is as follow:
F⁻(aq) + H₂O(l) ⇄ HF(aq) + OH⁻(aq)

3 0

3 years ago

Read 2 more answers

For a nucleus to go through mitosis,

ASHA 777 [7]

Answer:

A

Explanation:

I had this in a test

8 0

3 years ago

If given the mole fraction of water in an aqueous sucrose (C12H22O11) solution and asked to calculate the molality of the soluti

olasank [31]

Answer:

No additional information is required in order to calculate the molality of the solution.

Explanation:

Mole fraction of water = $\chi_1$

Mole fraction of sucrose = $\chi_2$

$\chi_1+\chi_2=1$

From the given mole fraction of water mole fraction of sucrose can be determined.

And from both values of mole fraction moles of water and sucrose can also be calculated.

$\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_2=\frac{n_2}{n_1+n_2}$

After calculating the moles of water convert them into mass of water in grams and then change that into kilograms say that be M'.

Finally we can determine the molality of solution ;

$Molality=\frac{\text{Moles of compound}}{\text{mass of solvent (kg)}}$

Molality of the sucrose solution ;

$m=\frac{n_1}{M'}$

From this we can conclude that no additional information is required in order to calculate the molality of the solution.

7 0

3 years ago

Question 1 of 6

Juli2301 [7.4K]

Answer:

Option A. 2, 3, 2

Explanation:

We'll begin by balancing the equation. This can be achieved by doing the following:

Fe + Cl2 —> FeCl3

There are 2 atoms of Cl on the left side and 3 atoms on the right side. It can be balance by putting 3 in front of Cl2 and 2 in front of FeCl3 as shown below:

Fe + 3Cl2 —> 2FeCl3

There are 2 atoms of Fe on the right side and 1 atom on the left side. It can be balance by putting 2 in front of Fe as shown below:

2Fe + 3Cl2 —> 2FeCl3

Now the equation is balanced.

The coefficients are : 2, 3, 2

8 0

3 years ago