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3 years ago

If the pH of an acid is 4.0, what is the pOH?

2 answers:

6 0

Well from my research I've calculated that pH + pOH = 14.00. in conclusion
pOH = 14.0 - 4.0 and so forth that equals 10.0 your welcome

My name is Ann [436]3 years ago

4 0

Well from research, I've calculated that pH + pOH = 14.00. In the end, pOH = 14.0 - 4.0 and so forth that equals 10.0.

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Moving water carrying away small

kondaur [170]

Answer:

erosion:)

Explanation:

7 0

3 years ago

Read 2 more answers

Addition of 50. J to a 10.0-g sample of a metal will cause the temperature of a metal to rise from 25ºC to 35ºC. The specific he

Snowcat [4.5K]

Answer:

b) C = 0.50 J/(g°C)

Explanation:

Q = mCΔT

∴ Q = 50 J

∴ m = 10.0 g

∴ ΔT = 35 - 25 = 10 °C

specific heat (C) :

⇒ C = Q / mΔT

⇒ C = 50 J / (10.0 g)(10 °C)

⇒ C = 0.50 J/(g°C)

7 0

3 years ago

A solution made by dissolving 33 mg of insulin in 6.5 mL of water has an osmotic pressure of 15.5 mmHg at 25°C. Calculate the mo

Liula [17]

<u>Answer:</u> The molar mass of the insulin is 6087.2 g/mol

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 15.5 mmHg

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (insulin) = 33 mg = 0.033 g (Conversion factor: 1 g = 1000 mg)

Volume of solution = 6.5 mL

R = Gas constant = $62.364\text{ L.mmHg }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$15.5mmHg=1\times \frac{0.033\times 1000}{\text{Molar mass of insulin}\times 6.5}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of insulin}=\frac{1\times 0.033\times 1000\times 62.364\times 298}{15.5\times 6.5}=6087.2g/mol$

Hence, the molar mass of the insulin is 6087.2 g/mol

8 0

3 years ago

58. Which of the following aqueous solutions has the lowest freezing point? A. 0.2 M NaCl B. 0.2 M FeCl3 C. 0.2M sucros D. 0.2 M

m_a_m_a [10]

The correct option is D. 0.2 M CaCl2 is has the lowest freezing point.

<h3>What is aqueous solution?</h3>

When one significance liquefies into another, a solution is formed. A solution is a homogenous mixture consisting of a solute dissolved into a solvent. The solute is the essence that is being dissolved, while the solvent is the dissolving medium. Solutions can be formed with multiple different classifications and forms of solutes and solvents. In this branch, we will focus on a resolution where the solvent is water.

An aqueous solution is a moisture that contains one or more dissolved essence. The dissolved importance in an aqueous solution may be solids, gases, or different liquids.
In directive to be a true solution, an assortment must be stable. When sugar is fully dissolved into moisture, it can stand for an undetermined amount of time, and the sugar will not recompense out of the solution. Further, if the sugar-water solution is passed through a filter, it will stay with the water.
This is because the liquefied particles in a resolution are very small, usually less than 1nm in diameter. Solute particles can be atoms, ions, or molecules, counting on the type of essence that has been dissolved.

To learn more about aqueous solution, refer to:

brainly.com/question/14469428

#SPJ9

6 0

2 years ago

Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base, and write the Ka expression for

Katyanochek1 [597]

Answer:

See explanation below

Explanation:

There are several ways to know if an acid or base is strong. One method is calculating the pH. If the pH is really low, is a strong acid, and if it's really high is a strong base.

However we do not have a pH value here.

The other method is using bronsted - lowry theory. If an acid is strong, then his conjugate base is weak. Same thing with the bases.

Now, Looking at the 4 compounds, we can say that only two of them is weak and the other two are strong compounds. Let's see:

LiOH ---> Strong. If you try to dissociate :

LiOH ------> Li⁺ + OH⁻ The Li⁺ is a weak conjugate acid.

HF -----> Weak

HF --------> H⁺ + F⁻ The Fluorine is a relatively strong conjugate base.

HCl -----> Strong

This is actually one of the strongest acid.

NH₃ ------> Weak

Now writting the Ka and Kb expressions:

Ka = [H⁺] [F⁻] / [HF]

Kb = [NH₄⁺] [OH⁻] / [NH₃]

Finally, to calculate the [OH⁻] we need to use the following expression:

Kw = [H⁻] [OH⁻]

Solving for [OH⁻] we have:

[OH⁻] = Kw / [H⁺]

Remember that the value of Kw is 1x10⁻¹⁴. So replacing:

[OH⁻] = 1x10⁻¹⁴ / 7x10⁻⁶

[OH⁻] = 1.43x10⁻⁹ M

And now, multiplying by 10¹⁰ we have:

[OH⁻] = 1.429x10⁻⁹ * 1x10¹⁰

Hope this helps

4 0

3 years ago