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iris [78.8K]
2 years ago
13

it takes 513 kj to remove a mole of electrons from the atoms at the surface of a piece of metal. how much energy does it take to

remove a single electron from n atom at the surface of the metal
Chemistry
1 answer:
schepotkina [342]2 years ago
8 0

Answer:

The right solution is "8.5\times 10^{-19} \ joule".

Explanation:

As we know,

1 mole electron = 6.023\times 10^{23} \ no. \ of \ electrons

Total energy = 513 \ KJ

                     = 513\times 1000 \ joule

For single electron,

The amount of energy will be:

= \frac{513\times 1000}{(6.023\times 10^{23})}

= 8.5\times 10^{-19} \ joule

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If a 0.10 M solution of a colored substance has a maximum absorbance at 500 nm and an absorbance of 0.26 M at this wavelength, w
Mashcka [7]
Absorbance is related to the concentration of a substance using the Beer-Lambert's Law. According to this law, absorbance is linearly related to concentration. However, this is only true up to a certain concentration depending on the substance. For this case, we assume that the said law is applicable.

A = kC
Using the first conditions, ewe solve for k.
0.26 = k (0.10)
k = 2.6

A = kC
A = 2.6 (0.20) = 0.52

Therefore, the absorbance at a concentration of 0.20 M and wavelength of 500nm is 0.52.
7 0
2 years ago
At which of the following temperatures is the average kinetic energy of the molecules of a substance the least? 23.2 °C 35.2 °C
svp [43]
At 15.2°C. Kinetic energy of molecules highly depends on the temperature — the warmer it is, the faster the molecules will move, especially in fluids (gases and liquids). If we consider that the formula for average kinetic energy of molecules is:

Ek = 3/2*k*T where k is Boltzmanns constant and 3/2 is, well, 3/2, kinetic energy of molecules really only depends on the temperature.
7 0
2 years ago
To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple _______. Then, plot ln(K
Aloiza [94]

Answer:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us use the thermodynamic definition of the Gibbs free energy and its relationship with Ksp as follows:

\Delta G=-RTln(Ksp)\\\\\Delta G=\Delta H-T\Delta S

Thus, by combining them, we obtain:

-RTln(Ksp)=\Delta H-T\Delta S\\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{T\Delta S}{RT} \\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{\Delta S}{R}

Which is related to the general line equation:

y=mx+b

Whereas:

y=ln(Ksp)\\\\m=-\frac{\Delta H}{R} \\\\x=\frac{1}{T} \\\\b=\frac{\Delta S}{R}

It means that we answer to the blanks as follows:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Regards!

8 0
2 years ago
Which of the following is consistent with a spontaneous process in the forward direction?
Kryger [21]

Answer:

A) ∆Suniv >0, ∆G<0, T∆Suniv >0.

Explanation:

The connection between entropy and the spontaneity of a reaction is expressed by the <u>second law of thermodynamics</u><u>: The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process</u>.

Mathematically, we can express the second law of thermodynamics as follows:

For a spontaneous process: ΔSuniv = ΔSsys + ΔSsurr > 0

Therefore, the second law of thermodynamics tells us that a spontaneous reaction increases the entropy of the universe; that is, ΔSuniv > 0.

If we want spontaneity expressed only in terms  of the properties of the system (ΔHsys and ΔSsys), we use the following equation:

-TΔSuniv = ΔHsys - TΔSsys < 0

That means that T∆Suniv >0.

This equation says that for a process carried out at constant pressure and temperature  T, if the changes in enthalpy and entropy of the system are such that <u>ΔHsys - TΔSsys  is less than zero, the process must be spontaneous.</u>

Finally, if the change in free energy is less than zero (ΔG<0), the reaction is spontaneous in the forward direction.

7 0
2 years ago
How many grams of ca(no3)2 can be produced by reacting excess hno3 with 6.33 g of ca(oh)2?
xeze [42]

Answer:

Amount of Ca(NO3)2 produced = 14.02 g

Explanation:

The given reaction can be depicted as follows:

Ca(OH)2 + 2HNO3 → Ca(NO3)2 + 2H2O

Since it is given that HNO3 is in excess, the limiting reactant is Ca(OH)2

Now, Mass of Ca(OH)2 = 6.33 g

Molar mass of Ca(OH)2 = 74 g/mol

Moles\ Ca(OH)2 = \frac{Mass}{Molar\ Mass} = \frac{6.33 g}{74 g/mol} =0.0855

Based on the reaction stoichiometry:

1 mole of Ca(OH)2 forms 1 mole of Ca(NO3)2

Therefore, moles of Ca(NO3)2 produced from the moles of Ca(OH)2 reacted = 0.0855 moles

Molar mass of Ca(NO3)2 = 164 g/mol

Mass\ Ca(NO3)2 \ produced = moles*molar\ mass \\= 0.0855\ moles*164\ g/mol = 14.02\  g

6 0
2 years ago
Read 2 more answers
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