If the pH of a solution is 8.45, what is the concentration of the OH-?

Chemistry

1 answer:

netineya [11]2 years ago

3 0

Answer:

How do you find the concentration of OH in a solution?

The hydroxide ion concentration can be found from the pOH by the reverse mathematical operation employed to find the pOH. [OH-] = 10-pOH or [OH-] = antilog ( - pOH) Page 2 Example: What is the hydroxide ion concentration in a solution that has a pOH of 5.70? On a calculator calculate 10-5.70, or "inverse" log (- 5.70).Explanation:

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ruslelena [56]

Answer:

The balanced equation is:

2 HNO3 + Mg ---> Mg(NO3)2 + H2

From the equation, we can see that we need twice the moles of HNO3 than the moles of Mg

Moles of Mg:

Molar mass of Mg = 24 g/mol

Moles = Given mass / Molar Mass

Moles of Mg = 4.47 / 24 = 0.18 moles (approx)

Hence, 2(moles of Mg) = 0.36 moles of HNO3 will be consumed

Number of moles of HNO3 after the reaction is finished is the number of unreacted moles of HNO3

Unreacted moles of HNO3 = Total Moles - Moles consumed

Unreacted moles of HNO3 = 0.64 moles (approx)

Since we approximated the value of moles of Mg, the value of remaining moles of HNO3 will also be approximate

From the given options, we can see that 0.632 moles is the closest value to our answer

Therefore, 0.632 moles will remain after the reaction

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3 years ago

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Nuetrik [128]

Answer: If a hydrogen atom and a helium atom have the same kinetic energy then the wavelength of the hydrogen atom will be roughly equal to the wavelength of the helium atom.

Explanation:

The relation between energy and wavelength is as follows.

$E = \frac{hc}{\lambda}\\$

This means that energy is inversely proportional to wavelength.

As it is given that energy of a hydrogen atom and a helium atom is same.

Let us assume that $E_{hydrogen} = E_{helium} = E'$ . Hence, relation between their wavelengths will be calculated as follows.

$E_{hydrogen} = \frac{hc}{\lambda_{hydrogen}}$ ... (1)

$E_{helium} = \frac{hc}{\lambda_{helium}}$ ... (2)

Equating the equations (1) and (2) as follows.

$E_{hydrogen} = E_{helium} = E'\\\frac{hc}{\lambda_{hydrogen}} = \frac{hc}{\lambda_{helium}} = E'\\\lambda_{helium} = \lambda_{hydrogen} = E'$

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irinina [24]

Answer:

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