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Charra [1.4K]
2 years ago
13

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

#SPJ4

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1.73 M

Explanation:

Molarity is moles per liter, so we need to divide 7.47 moles by 4.31 liters to get the molarity of the solution.

7.47/4.31 ≈ 1.73 M

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If gas is initially at 350L and 500k then changes to 295K what is the new volume
Dimas [21]
<h2>Hello!</h2>

The answer is:

The new volume is equal to 206.5 L.

<h2>Why?</h2>

To solve this problem, we need to assume that the pressure is constant, and use the Charle's Law equation, so, solving we have:

\frac{V_1}{T_1}=\frac{V_2}{T_2}

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Then, using the Charle's Law equation, we have:

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Have a nice day!

5 0
3 years ago
Find the volume of hydrogen gas formed when 1.5g of aluminum reacts with aq NaOH at 27 degrees Celcius.
Rudiy27

Answer:

how can I solve this ?4Al+3O2 produce 2Al2O3 find a) oxygen atoms needed to react with 5.4 g of aluminium b) grams of oxygen needed to react with 0.6 mol of aluminium?

(A) n=m/M,

n(Al)=5.4/27=0.2 moles

n(O2)=n(Al)*3/4=0.2*3/4=0.15 moles

Number of oxygen atoms= n(O2)*Avogadro's number

=0.15*6.02*10^23=9.03*10^22 oxgyen atoms

(B)

n=m/M

n(Al)=0.6/27=0.02222 moles

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8 0
2 years ago
For the reaction o(g) + o2(g) → o3(g) δh o = −107.2 kj/mol given that the bond enthalpy in o2(g) is 498.7 kj/mol, calculate the
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The reaction;
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ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
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=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
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