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ICE Princess25 [194]
2 years ago
5

Use coulomb's law to calculate the ionization energy in kj/mol of an atom composed of a proton and an electron separated by 185.

00 pm .
Chemistry
2 answers:
xz_007 [3.2K]2 years ago
6 0

Answer:

Energy = 752.88 kJ/mol

Explanation:

The force of attraction can be written as:

F=\frac {K\times |Z_{cation}|e\times |Z_{anion}|e}{r^2}

Integrating to find the expression for energy. We get that:-

E=\frac {K\times |Z_{cation}|e\times |Z_{anion}|e}{r}

Where,  

K is the Coulomb's constant having value 9×10⁹ N. m²/C²

Z_{cation}  is the charge on the cation

Z_{anion}  is the charge on the anion

e is electronic charge = Charge of the electron = Charge on proton 1.602\times 10^{-19}\ C

r is the distance between the cation and anion  = 185.00 pm

Also, 1 pm = 10⁻¹² m

So, r = 185×10⁻¹² m

Z_{cation}=Z_{anion}=1

Applying in the equation,  

E=\frac{9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{185\times 10^{-12}}=1.25\times 10^{-18}\ J/atom

Also,

1 atom = 6.023\times 10^{23}\ mole^{-1}

So,

Energy = 1.25\times 10^{-18}\times 6.023\times 10^{23}\ J/mole=752875\ J/mole

Also, 1 J = 0.001 kJ

So, <u>Energy = 752.88 kJ/mol</u>

Tems11 [23]2 years ago
5 0
Coulomb's law mathematically is:
F = kQ₁Q₂/r²
we integrate this with respect to distance to obtain the expression for energy:
E = kQ₁Q₂/r; where k is the Coulomb's constant = 9 x 10⁹; Q are the charges, r is the seperation
Charge on proton = charge on electron = 1.6 x 10⁻¹⁹ C
E = (9 x 10⁹ x 1.6 x 10⁻¹⁹ x 1.6 x 10⁻¹⁹) / (185 x 10⁻¹²)
E = 1.24 x 10⁻¹⁸ Joules per proton/electron pair
Number of pairs in one mole = 6.02 x 10²³
Energy = 6.02 x 10²³ x 1.24 x 10⁻¹⁸
= 746.5 kJ
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calculate the water potential of a solution of 0.15m sucrose. the solution is at standard temperature.
Mrac [35]

Answer:

The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.

Explanation:

Water potential = Pressure potential + solute potential

P_w=P_p+P_s

P_w=P_p+(-iCRT)

We have :

C = 0.15 M, T = 273.15 K

i = 1

The water potential of a solution of 0.15 m sucrose= P_w

P_p=0 bar (At standard temperature)

P_s=-iCRT=-\times 1\times 8.314\times 10^{-2}bar L/mol K\times 273.15 K=-3.406 bar

P_w=0 bar+(-3.406 ) bar

The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.

7 0
2 years ago
Assuming ideal solution behavior, what is the freezing point of a solution of 9.04 g of I2 in 75.5 g of benzene?
cluponka [151]

Answer:

3.1°C

Explanation:

Using freezing point depression expression:

ΔT = Kf×m×i

<em>Where ΔT is change in freezing point, Kf is freezing point depression constant (5.12°c×m⁻¹), m is molality of the solution and i is Van't Hoff factor constant (1 For I₂ because doesn't dissociate in benzene).</em>

Molality of 9.04g I₂ (Molar mass: 253.8g/mol) in 75.5g of benzene (0.0755kg) is:

9.04g ₓ (1mol / 253.8g) = 0.0356mol I₂ / 0.0755kg = 0.472m

Replacing in freezing point depression formula:

ΔT = 5.12°cm⁻¹×0.472m×1

ΔT = 2.4°C

As freezing point of benzene is 5.5°C, the new freezing point of the solution is:

5.5°C - 2.4°C =

<h3>3.1°C</h3>

<em />

3 0
2 years ago
The three‑dimensional structure of a generic molecule is given. Identify the axial and equatorial atoms in the three‑dimensional
Dima020 [189]

Answer:

Explanation:

CHECK THE ATTACHMENT FOR THE COMPLETE QUESTION AND THE DETAILED EXPLANATION

NOTE:

Equatorial atoms are referred to atoms that are attached to carbons in the cyclohexane ring which is found at the equator of the ring.

Axial atoms are atoms that exist in a bond which is parallel to the axis of the ring in cyclohexane

4 0
2 years ago
During a titration the following data were collected. A 10. mL portion of an unknown monoprotic acid solution was titrated with
Liula [17]

Answer:

8.0 moles

Explanation:

Since the acid is monoprotic, 1 mole of the acid will be required to stochiometrically react with 1 mole of NaOH.

Using the formula: \frac{concentration of acid X volume of acid}{concentration of base X volume of base} = \frac{mole of acid}{mole of base}

Concentration of acid = ?

Volume of acid = 10 mL

Concentration of base = 1.0 M

Volume of base = 40 mL

mole of acid = 1

mole of base = 1

Substitute into the equation:

\frac{concentration of acid X 10}{1.0 X 40} = \frac{1}{1}

Concentration of acid = 40/10 = 4.0 M

To determine the number of moles of acid present in 2.0 liters of the unknown solution:

Number of moles = Molarity x volume

molarity = 4.0 M

Volume = 2.0 Liters

Hence,

Number of moles = 4.0 x 2.0 = 8 moles

8 0
2 years ago
3. According to the label on a bottle of concentrated hydrochloric acid, the contents are 36.0% HCl by mass and have a density o
velikii [3]

Answer:

a) 11.64 M

b) 43 mL

c) 1.7 kg

Explanation:

a) Let's use a basis of the calculus of 1000 mL (1 L) of the concentrated solution. If the solution has 1.18 g/mL, it has:

1.18*1000 = 1180 g.

The mass of HCl will be then:

mHCl = 1180*0.36 = 424.8 g

The molar mass of HCl is 36.5 g/mol, so the number of moles is the mass divided by the molar mass:

nHCl = 424.8/36.5 = 11.64 mol

The molarity is the number of moles divided by the volume in L:

Molarity = 11.64 M

b) To prepare a solution by dilution of a concentrated one, we can use the equation:

C1V1 = C2V2

Where C is the concentration, V is the volume, 1 is the concentrated solution, and 2 the final solution. So:

11.64*V1 = 2.00*0.250

V1 = 0.0429 L ≅ 43 mL

c) The neutralization will happen by the equation:

HCl + NaHCO₃ → NaCl + CO₂ + H₂O

So, 1 mol of NaHCO₃ is needed to react with 1 mol of HCl. At 1.75 L, the number of moles of the acid is:

nHCl = 1.75*11.64 = 20.37 mol

The molar mass of NaHCO₃ is 84 g/mol so the mass needed is the molar mass multiplied by the number of moles:

m = 84*20.37 = 1,711.08 g

m = 1.7 kg

6 0
2 years ago
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