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3 years ago

Which characteristic of an atom always varies among atoms of different elements

Chemistry

1 answer:

Alex787 [66]3 years ago

6 0

The number of protons in the nucleus always varies among atoms of different elements.

Every element has a different number of protons than every other element.

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The small bags of silica gel you often see in a new shoe box are placed there to control humidity. Despite its name, silica gel

Zinaida [17]

So we look equation for the free Gibbs free energy (ΔG) which depends on entalpy (ΔH), temperature (T) and entropy (ΔS):

ΔG = ΔH - TΔS

ΔG is negative (-) because the water absorption on the silica gel surface is a spontaneous process.

ΔH is negative (-) because the water absorption on the silica gel surface is a exothermic process (it releases heat and if you want to desorb the water form the silica gen you need to add heat which is a endothermic process).

ΔS is negative (-) because the water is adsorbed, so from disorderly state you take the water molecules and put them in a orderly state and by doing that you decrease the entropy.

6 0

3 years ago

Gold forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimeter for a silver-gold

Mama L [17]

<u>Answer:</u> The number of gold atoms per cubic centimeters in the given alloy is $1.83\times 10^{22}$

<u>Explanation:</u>

To calculate the number of gold atoms per cubic centimeters for te given silver-gold alloy, we use the equation:

$N_{Au}=\frac{N_AC_{Au}}{(\frac{C_{Au}M_{Au}}{\rho_{Au}})+(\frac{M_{Au}(100-C_{Au})}{\rho_{Ag}})}$

where,

$N_{Au}$ = number of gold atoms per cubic centimeters

$N_A$ = Avogadro's number = $6.022\times 10^{23}atoms/mol$

$C_{Au}$ = Mass percent of gold in the alloy = 42 %

$\rho_{Au}$ = Density of pure gold = $19.32g/cm^3$

$\rho_{Ag}$ = Density of pure silver = $10.49g/cm^3$

$M_{Au$ = molar mass of gold = 196.97 g/mol

Putting values in above equation, we get:

$N_{Au}=\frac{(6.022\times 10^{23}atoms/mol)\times 48\%}{(\frac{48\%\times 196.97g/mol}{19.32g/cm^3})+(\frac{196.97g/mol\times 58\%}{10.49g/cm^3})}\\\\N_{Au}=1.83\times 10^{22}atoms/cm^3$

Hence, the number of gold atoms per cubic centimeters in the given alloy is $1.83\times 10^{22}$

5 0

3 years ago

What bond vibrations and stretching frequencies should you observe in the IR spectrum of caffeine?

Margarita [4]

The bond vibrations that can be observed in the IR spectrum of caffeine :

i) asymmetric, ii) symmetric, iii) wagging, iv) twisting, v) scissoring, vi) rocking

while the observable stretching frequencies are ; 1700, 1300, 2900, 1500 and 3000.

Caffeine is a polyatomic molecule ( contains more than two atoms bonded via covalent bonds ) therefore its atoms can vibrate in three dimensions ( x, y, z ).therefore the bond vibrations that can be observed in the IR spectrum of caffeine are : asymmetric, symmetric, wagging, twisting, scissoring, and rocking.

The observable stretching frequencies that are in the IR spectrum of caffeine ; C = O is 1700 , C - H is 1300, C -H is 2900, C -N is 1500, and N -H = 3000.

Hence we can conclude The bond vibrations that can be observed in the IR spectrum of caffeine : i) asymmetric, ii) symmetric, iii) wagging, iv) twisting, v) scissoring, vi) rocking and the observable stretching frequencies are ; 1700, 1300, 2900, 1500 and 3000.

Learn more : brainly.com/question/13184210

3 0

2 years ago

Use the data given below to construct a Born-Haber cycle to determine the second ionization energy of Ca. Δ H°(kJ) Ca(s)→Ca(g) 1

Drupady [299]

Answer : The value of second ionization energy of Ca is 1010 kJ.

Explanation :

The formation of calcium oxide is,

$Ca(s)+\frac{1}{2}O_2(g)\overset{\Delta H_f}\rightarrow CaO(s)$

$\Delta H_f^o$ = enthalpy of formation of calcium oxide = -635 kJ

The steps involved in the born-Haber cycle for the formation of $CaO$ :

(1) Conversion of solid calcium into gaseous calcium atoms.

$Ca(s)\overset{\Delta H_s}\rightarrow Ca(g)$

$\Delta H_s$ = sublimation energy of calcium = 193 kJ

(2) Conversion of gaseous calcium atoms into gaseous calcium ions.

$Ca(g)\overset{\Delta H_I_1}\rightarrow Ca^{+1}(g)$

$\Delta H_I_1$ = first ionization energy of calcium = 590 kJ

(3) Conversion of gaseous calcium ion into gaseous calcium ions.

$Ca^{+1}(g)\overset{\Delta H_I_2}\rightarrow Ca^{+2}(g)$

$\Delta H_I_2$ = second ionization energy of calcium = ?

(4) Conversion of molecular gaseous oxygen into gaseous oxygen atoms.

$O_2(g)\overset{\Delta H_D}\rightarrow OI(g)$

$\frac{1}{2}O_2(g)\overset{\Delta H_D}\rightarrow O(g)$

$\Delta H_D$ = dissociation energy of oxygen = $\frac{498}{2}=249kJ$

(5) Conversion of gaseous oxygen atoms into gaseous oxygen ions.

$O(g)\overset{\Delta H_E_1}\rightarrow O^-(g)$

$\Delta H_E_1$ = first electron affinity energy of oxygen = -141 kJ

(6) Conversion of gaseous oxygen ion into gaseous oxygen ions.

$O^-(g)\overset{\Delta H_E_2}\rightarrow O^{2-}(g)$

$\Delta H_E_2$ = second electron affinity energy of oxygen = 878 kJ

(7) Conversion of gaseous cations and gaseous anion into solid calcium oxide.

$Ca^{2+}(g)+O^{2-}(g)\overset{\Delta H_L}\rightarrow CaO(s)$

$\Delta H_L$ = lattice energy of calcium oxide = -3414 kJ

To calculate the overall energy from the born-Haber cycle, the equation used will be:

$\Delta H_f^o=\Delta H_s+\Delta H_I_1+\Delta H_I_2+\Delta H_D+\Delta H_E_1+\Delta H_E_2+\Delta H_L$

Now put all the given values in this equation, we get:

$-635kJ=193kJ+590kJ+\Delta H_I_2+249kJ+(-141kJ)+878kJ+(-3414kJ)$

$\Delta H_I_2=1010kJ$

Therefore, the value of second ionization energy of Ca is 1010 kJ.

6 0

3 years ago

What is the pH at the half-equivalence point in the titration of a weak base with a strong acid? The pKb of the weak base is 7.9

ValentinkaMS [17]

Hey there!

Given the reaction:

B + H⁺ => HB⁺

At half-equivalence point : [B] = [HB⁺]

=> [B] / [HB⁺] = 1

Henderson-Hasselbalch equation :

pH = pKa + log ( [B] ) / ( HB⁺)]

pH = 14 - pKb + log ( 1 )

pH = 14 - 7.95 + 0

pH = 6.05

Answer C

Hope that helps!

5 0

3 years ago