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3 years ago

The pOH of a solution of KOH is 11.30. What is the [H*] for this solution?

Chemistry

1 answer:

Sloan [31]3 years ago

4 0

Answer: The concentration of hydrogen ions for this solution is $1.99 \times 10^{-3}$ .

Explanation:

Given: pOH = 11.30

The relation between pH and pOH is as follows.

pH + pOH = 14

pH + 11.30 = 14

pH = 14 - 11.30

= 2.7

Also, pH is the negative logarithm of concentration of hydrogen ions.

$pH = - log [H^{+}]$

Substitute the values into above formula as follows.

$pH = -log [H^{+}]\\2.7 = -log [H^{+}]\\conc. of H^{+} = 1.99 \times 10^{-3}$

Thus, we can conclude that the concentration of hydrogen ions for this solution is $1.99 \times 10^{-3}$ .

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2 years ago

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sveticcg [70]

Answer:

12.36 L.

Explanation:

We'll begin by calculating the number of mole in 147.1 g of lead(II) nitrate, Pb(NO₃)₂. This can be obtained as follow:

Molar mass of Pb(NO₃)₂ = 207.2 + 2[14.01 + (16×3)]

= 207.2 + 2[14.01 + 48]

= 207.2 + 2[62.01]

= 207.2 + 124.02

= 331.22 g/mol

Mass of Pb(NO₃)₂ = 147.1 g

Mole of Pb(NO₃)₂ =?

Mole = mass / Molar mass

Mole of Pb(NO₃)₂ = 147.1 / 331.22

Mole of Pb(NO₃)₂ = 1.104 moles.

Next, we shall determine the number of mole of oxygen gas, O₂, produce from the reaction. This can be obtained as follow:

2Pb(NO₃)₂ —> 2PbO + 4NO₂ + O₂

From the balanced equation above,

2 moles of Pb(NO₃)₂ decomposed to produce 1 mole of O₂.

Therefore, 1.104 moles of Pb(NO₃)₂ will decompose to produce = (1.104 × 1)/2 = 0.552 mole of O₂.

Finally, we shall determine the volume occupied by 0.552 mole of oxygen gas, O₂. This can be obtained as follow:

1 mole of O₂ occupied 22.4 L at STP.

Therefore, 0.552 mole of O₂ will occupy = 0.552 × 22.4 = 12.36 L at STP.

Thus, the volume of oxygen gas, O₂ produced is 12.36 L.

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2 years ago

A 50.0 g sample of scandium, sc, is heated by exposure to 1.50 x 10 3 j. The temperature of the sc is raised by 61.1 o

statuscvo [17]

Given mass of Scandium = 50.0 g

Increase in temperature of the metal when heated = $61.1^{0}C$

Heat absorbed by Scandium = $1.50*10^{3}J$

The equation showing the relationship between heat, mass, specific heat and temperature change:

$Q = m C (deltaT)$

Where Q is heat = $1.50*10^{3}J$

m is mass = 50.0 g

ΔT = $61.1^{0}C$

On plugging in the values and solving for C(specific heat) we get,

$1.50*10^{3}J$ =50.0g(C)( $61.1^{0}C$ )

C = 0.491 $\frac{J}{g^{0}C }$

Specific heat of the metal = 0.491 $\frac{J}{g^{0}C }$

7 0

3 years ago

The pOH of a solution of KOH is 11.30. What is the [H*] for this solution?​

The pOH of a solution of KOH is 11.30. What is the [H*] for this solution?