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2 years ago

Calculate the concentration (in m) of hydronium ions in a solution at 25. 0 °C with a POH of 3. 58.

Chemistry

1 answer:

Artyom0805 [142]2 years ago

8 0

The concentration (in M) of hydronium ions in a solution at 25. 0 °C with a POH of 3. 58 Is 3.8× 10⁻¹¹ M

pOH is the measure of basic nature of a solution by evaluating the [OH⁻] concentration.

pH is measure of acidic nature of a solution by evaluating the [H⁺] concentration. It is the negative logarithm of the hydroxide ion concentration. It gives hydronium ions on dissociation.

Given,

pOH = 3.58

Temperature = 25°C = 298K

We know that pH + pOH =14

pH + 3.58 = 14

pH = 10.42

pH is the negative logarithm of hydrogen ion concentration.

At 25°C, the relation of pH and [H⁺] concentration is as follows:

∴ pH = -log [H⁺]

⇒ 10.42 = -log [H⁺]

⇒log [H⁺] = -10.42

⇒ [H⁺] = antilog (-10.42)

⇒[H⁺] =3.8× 10⁻¹¹ M

The concentration (in m) of hydronium ions in a solution at 25. 0 °C with a POH of 3. 58 Is 3.8× 10⁻¹¹ M

Learn more about pH here, brainly.com/question/17144456

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3 years ago

Can you discount alcohol in Texas?.

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1 year ago

To balance a chemical equation it may be necessary to adjust the

Mars2501 [29]

Answer:

<em>To balance a chemical equation it may be necessary to adjust the </em><u>coefficients.</u>

Explanation:

The <em>coefficients</em> of a <em>chemical equation</em> are the numbers that you put in front of each reactant and product. They are used to balance the equation and comply with the law of mass conservation.

By adjusting the coefficients you obtain the relative amounts (moles) of each product and reactant, i.e. the mole ratios.

Here an example.

The first information is what is called a word equation. E.g. nitrogen and hydrogen react to form ammonia:

Word equation: hydrogen + nitrogen → ammonia

Skeleton equation: H₂ + N₂ → NH₃

This equation shows the chemical formulae but it is not balanced. The law of mass conservation is not observed.

So, in order to comply with the law of mass conservation you adjust the coefficients as follow.

Balanced chemical equation: 3H₂ + N₂ → 2NH₃

As you see, it was necessary to modify the coefficients. Now the law of conservation of mass is observed and you get the mole ratios:

3 mol H₂ : 1 mol N₂ : 2 mol NH₃

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3 years ago

a 125 g chunk of aluminum at 182 degrees Celsius was added to a bucket filled with 365 g of water at 22.0 degrees Celsius. Ignor

Diano4ka-milaya [45]

<h3>Answer:</h3>

32.98°C

<h3>Explanation:</h3>

We are given the following;

Mass of Aluminium as 125 g

Initial temperature of Aluminium as 182°C

Mass of water as 265 g

Initial temperature of water as 22°C

We are required to calculate the final temperature of the two compounds;

First, we need to know the specific heat capacity of each;

Specific heat capacity of Aluminium is 0.9 J/g°C

Specific heat capacity of water is 4.184 J/g°C

<h3>Step 1: Calculate the Quantity of heat gained by water.</h3>

Assuming the final temperature is X°C

we know, Q = mcΔT

Change in temperature, ΔT = (X-22)°C

therefore;

Q = 365 g × 4.184 J/g°C × (X-22)°C

= (1527.16X-33,597.52) Joules

<h3>Step 2: Calculate the quantity of heat released by Aluminium </h3>

Using the final temperature, X°C

Change in temperature, ΔT = -(X°- 182°)C (negative because heat was lost)

Therefore;

Q = 125 g × 0.90 J/g°C × (182°-X°)C

= (20,475- 112.5X) Joules

<h3>Step 3: Calculating the final temperature</h3>

We need to know that the heat released by aluminium is equal to heat absorbed by water.

Therefore;

(20,475- 112.5X) Joules = (1527.16X-33,597.52) Joules

Combining the like terms;

1639.66X = 54072.52

X = 32.978°C

= 32.98°C

Therefore, the final temperature of the two compounds will be 32.98°C

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4 years ago

A sample of oxygen gas occupies 3.60 liters at a pressure of 1.00 atm. If temperature is held constant, what will be the volume

Soloha48 [4]

Answer: The volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L

At constant temperature, the volume of a fixed mass of gas is inversely proportional to the pressure it exerts, then

PV = c

Thus, if the pressure increases, the volume decreases, and if the pressure decreases, the volume increases.

It is not necessary to know the exact value of the constant c to be able to use this law since for a fixed amount of gas at constant temperature, it is satisfied that,

P₁V₁ = P₂V₂

Where P₁ and P₂ as well as V₁ and V₂ correspond to pressures and volumes for two different states of the gas in question.

In this case the first oxygen gas state corresponds to P₁ = 1.00 atm and V₁ = 3.60 L while the second state would be P₂ = 2.50 atm and V₂ = y. Substituting in the previous equation,

1.00 atm x 3.60 L = 2.50 atm x y

We cleared y to find V₂,

V₂ = y = $\frac{1.00 atm x 3.60 L}{2.50 atm}$ = 1.44 L

Then, <u>the volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L</u>

3 0

3 years ago