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3 years ago

What is the pH of a solution within a solution with pH = 4.50? [H +] = 3.25×10-6 M?

Chemistry

1 answer:

Lisa [10]3 years ago

6 0

Answer: 5.48

Explanation:

pH is the negative logarithm of hydrogen ion concentration in a solution.

Mathematically, pH = - log(H+)

where H+ represent the concentration of hydrogen ion

So, to get the pH of the solution with [H +] = 3.25×10-6 M:

Apply, pH = -log(H+)

pH = - log (3.25×10-6 M)

pH = - ( -5.48)

(Note that the minus signs will cancel out each other)

Therefore pH = 5.48

Now we know that the pH of the solution with hydrogen ion concentration of 3.25×10-6 M is 5.48 (i.e slightly acidic)

Thus, we can finally say 5.48 is the pH of the solution within a solution with pH = 4.50

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3 years ago

If an old car burns 0.05 mL oil every mile, How much oil will it burn if driven 100,000 miles? Show your answer in mL and L?.

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3 years ago

A chemist must prepare of hydrochloric acid solution with a pH of at . He will do this in three steps: Fill a volumetric flask a

Eva8 [605]

Answer:

1.7 mL

Explanation:

A chemist must prepare 550.0 mL of hydrochloric acid solution with a pH of 1.60 at 25 °C. He will do this in three steps: Fill a 550.0 mL volumetric flask about halfway with distilled water. Measure out a small volume of concentrated (8.0 M) stock hydrochloric acid solution and add it to the flask. Fill the flask to the mark with distilled water. Calculate the volume of concentrated hydrochloric acid that the chemist must measure out in the second step. Round your answer to 2 significant digits.

Step 1: Calculate [H⁺] in the dilute solution

We will use the following expresion.

pH = -log [H⁺]

[H⁺] = antilog - pH = antilog -1.60 = 0.0251 M

Since HCl is a strong monoprotic acid, the concentration of HCl in the dilute solution is 0.0251 M.

Step 2: Calculate the volume of the concentrated HCl solution

We want to prepare 550.0 mL of a 0.0251 M HCl solution. We can calculate the volume of the 8.0 M solution using the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂/C₁

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3 0

3 years ago

Please help me like now please

777dan777 [17]

Answer:

1-1) NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2

1-2) 0.5 mole of CO2

2-1) 2C4H10 + 13O2 --> 8CO2 + 10H2O

2-2) 4 mol CO2

Explanation:

Question 1

NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2

To balance the equation, count the number of atoms on both sides of the equation

(1 Na, 1+3+1H, 1+1+1C, 3+2Oxygen) --> (1 Na, 1+1+1C, 3+2H, 2+1+2Oxygen)

Combining the pluses will give you the following

(1 Na, 5H, 3C, 5Oxygen) --> (1 Na, 3C, 5H, 5Oxygen)

Both sides are the same, therefore the chemical equation is balanced (originally).

From the equation, we can see that 1 mole of NaHCO3 produces 1 mole of CO2.

So that means 0.5 mole of NaHCO3 would produce 0.5 mole of CO2.

Question 2

C4H10 + O2 --> CO2 + H2O

Again, count the number of atoms on both sides of the equation

(4C, 10H, 2O) --> (1C, 2H, 3O) This time left does not equal right side

You now need to find factors that can make both sides equal.

C4H10 + O2 --> 4CO2 + H2O Now the C is balanced, let's recount 

(4C, 10H, 2Oxygen) --> (4C, 8+1Oxygen, 2H) H&O is still not balanced

C4H10 + O2 --> 4CO2 + 5H2O Now the H is balanced, let's recount

(4C, 10H, 2Oxygen) --> (4C, 8+5Oxygen, 10H) O is still not balanced

C4H10 + (13/2)O2 --> 4CO2 + 5H2O Now the O is balanced

(4C, 10H, 13Oxygen) --> (4C, 13Oxygen, 10H)

But because 13/2 is a fraction, we want to eliminate that by multiplying every reactant and product by 2 (the denominator).

2C4H10 + 13O2 --> 8CO2 + 10H2O Now it's completely balanced!

(8C, 20H, 28Oxygen) --> (8C, 28Oxygen, 20H) Yayy! It's balanced.

Now, 2 mol C4H10 produces 8 mol CO2.

So 1 mol C4H10 produces 4 mol CO2.

6 0

3 years ago

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vovikov84 [41]

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4 years ago