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Marysya12 [62]
1 year ago
13

What laboratory conditions (temperature and pressure) were assumed when calculating ksp based on measured electrochemical potent

ial?
Chemistry
1 answer:
laiz [17]1 year ago
8 0

The temperature of the lab is assumed at 25°C while calculating Ksp. Pressure of the lab is assumed at 1 atm normal atmospheric conditions while calculating Ksp.

If we increase the temperature of the solvent, the average kinetic energies of the molecules start increases. Hence, the solvent is more able to dislodge more molecules from the surface of the solute. Thus, by increasing the temperature the solubilities of substances also increases. Temperature is directly proportional to the solubility.

As we goes on increasing the pressure of a gas, the collision between the molecules start increasing and thus the solubility goes up, as we decrease the pressure, the number of collision decreases thus solubility goes down.

Thus, with further increase in the temperature and pressure the Ksp value changes. Thus, temperature of the lab is assumed at 25°C while calculating Ksp. Pressure of the lab is must be 1 atm normal atmospheric conditions while calculating Ksp.

learn more about solubility:

brainly.com/question/14366471

#SPJ4

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shepuryov [24]

Answer:

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Explanation:

3 0
2 years ago
calculate the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom?
inessss [21]

<u>Answer:</u>

\Delta E=E_{final}-E_{initial}

\Delta E=-1312[\frac{1}{(n_f^2)}-\frac {1}{(n_i^2 )}]KJ mol^{-1}

\Delta E=-1312[\frac{1}{3^2)}-\frac {1}{(1^2 )}]KJ mol^{-1}

\Delta E=-1312[\frac{1}{(9)}-\frac {1}{(1 )}]KJ mol^{-1}

\Delta E=-1312[0.111-1]KJ mol^{-1}

\Delta E=1166 KJ mol^{-1}

\frac{=1166,000 \mathrm{J}}{6.022 \times 10^{23} \text { photons }}

=193623 \times 10^{-23}  \frac {J}{photon}

\Delta E=1.93623 \times 10^{-18}  \frac {J}{photon}

\Delta E=\frac {h\times c}{\lambda} \\\\=\frac {(6.626\times 10^{-34} J s \times 3 \times 10^8 ms^{-1})}{\lambda}

h is planck's constant  

c is the speed of light

λ is the wavelength of light  

\lambda =\frac {h\times c}{\Delta E}\\\\=\frac {(6.626\times10^{-34} J s\times3 \times 10^8 ms^{-1})}{(1.93623\times10^{-18}  J/photon)}

Wavelength

\lambda =10.3 \times 10^{-8} m \times \frac {(10^9 nm)}{1m}  =103 nm (Answer)

<em>Thus, the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom is </em><u><em>103 nm.</em></u>

7 0
3 years ago
How many grams are 1.25 moles of potassium bromide (KBr)?
sukhopar [10]

Answer:

The SI base unit for amount of substance is the mole. one mole is equal to one mole potassium bromide or 119.0023 grams

6 0
3 years ago
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stellarik [79]

Answer:

Australia is correct

Explanation:

Because it is not attached to any other place

3 0
3 years ago
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What is the correct name for SiCl5?
lyudmila [28]

Answer:

SbCl5

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7 0
3 years ago
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