Chem 211

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problem 1: relative magnitude of coulombic potential a. we are comparing a +2 ion separated by 10 nm from a +2 charge to a +3 ion separated by 5 nm from a +1 ion. coulomb's law states:

pe = k \frac{q _{1} q _{2}}{r}

, where

k = 9 \times 1 0^{9} n m^{2} / c^{2}

. for the first case:

q_{1} = + 2 c

,

q_{2} = + 2 c

,

r = 10 \times 1 0^{- 9} m

.

p e_{1} = (9 \times 1 0^{9}) \frac{( 2 \times 2 )}{10 \times 1 0 ^{- 9}} = 3.6 \times 1 0^{- 15} j

. for the second case:

q_{1} = + 3 c

,

q_{2} = + 1 c

,

r = 5 \times 1 0^{- 9} m

.

p e_{2} = (9 \times 1 0^{9}) \frac{( 3 \times 1 )}{5 \times 1 0 ^{- 9}} = 5.4 \times 1 0^{- 15} j

. the ratio of the potentials is

p e_{1} / p e_{2} = 3.6 \times 1 0^{- 15} /5.4 \times 1 0^{- 15} = 0.6667

, which is approximately

\frac{2}{3}

or 2:3. so, the coulombic potential of the +2 ion separated by 10 nm from a +2 charge is about 2/3 the magnitude of the coulombic potential of the +3 ion separated by 5 nm from a +1 ion. b. now, let's compare the a-3 ion separated by 6 nm from a +1 charge to a -2 ion separated by 5 nm from a +2 charge. for the first case:

q_{1} = - 3 c

,

q_{2} = + 1 c

,

r = 6 \times 1 0^{- 9} m

.

p e_{3} = (9 \times 1 0^{9}) \frac{( - 3 \times 1 )}{6 \times 1 0 ^{- 9}} = - 1.5 \times 1 0^{- 14} j

. for the second case:

q_{1} = - 2 c

,

q_{2} = + 2 c

,

r = 5 \times 1 0^{- 9} m

.

p e_{4} = (9 \times 1 0^{9}) \frac{( - 2 \times 2 )}{5 \times 1 0 ^{- 9}} = - 7.2 \times 1 0^{- 14} j

. the ratio of the potentials is

p e_{3} / p e_{4} = - 1.5 \times 1 0^{- 14} / - 7.2 \times 1 0^{- 14} = 0.2083

, which is approximately

\frac{1}{5}

or 1:5. so, the coulombic potential of the a-3 ion separated by 6 nm from a +1 charge is about 1/5 the magnitude of the coulombic potential of the -2 ion separated by 5 nm from a +2 charge. c. finally, let's compare the +2 ion separated by 10 nm from a +2 charge to a +3 ion separated by 20 nm from a +2 charge. for the first case:

q_{1} = + 2 c

,

q_{2} = + 2 c

,

r = 10 \times 1 0^{- 9} m

.

p e_{5} = (9 \times 1 0^{9}) \frac{( 2 \times 2 )}{10 \times 1 0 ^{- 9}} = 3.6 \times 1 0^{- 15} j

. for the second case:

q_{1} = + 3 c

,

q_{2} = + 2 c

,

r = 20 \times 1 0^{- 9} m

.

p e_{6} = (9 \times 1 0^{9}) \frac{( 3 \times 2 )}{20 \times 1 0 ^{- 9}} = 2.7 \times 1 0^{- 15} j

. the ratio of the potentials is

p e_{5} / p e_{6} = 3.6 \times 1 0^{- 15} /2.7 \times 1 0^{- 15} = 1.3333

, which is approximately

\frac{4}{3}

or 4:3. so, the coulombic potential of the +2 ion separated by 10 nm from a +2 charge is about 4/3 the magnitude of the coulombic potential of the +3 ion separated by 20 nm from a +2 charge. problem 2: volume of diamond given: density of diamond = 3.51 g/cm³, 1 carat = 200 mg. first, convert the carat weight to grams: 1 carat = 200 mg = 0.2 g next, use the formula for density: density = mass / volume volume = mass / density for a 1.50 carat diamond: mass = 1.50 carats * 0.2 g/carat = 0.30 g density = 3.51 g/cm³ volume = 0.30 g / 3.51 g/cm³ ≈ 0.0854 cm³ problem 3: energy released when bringing proton and electron close the energy released when a proton is brought from an infinite distance to within 53 pm of an electron can be calculated using coulomb's law:

e = k \frac{q _{1} q _{2}}{r}

where

k = 9 \times 1 0^{9} n m^{2} / c^{2}

,

q_{1}

is the charge of the proton (+1.602 x 10^-19 c),

q_{2}

is the charge of the electron (-1.602 x 10^-19 c), and

r = 53 \times 1 0^{- 12} m

. ( e = (9 imes 10^9) \frac{(1.602 imes 10^{-19} imes -1.602question 4: picket fence dimensional analysis to calculate the total volume of water corresponding to 2.00 inches of rainfall across 1.00 square mile, you can use the following steps: step 1: convert inches to feet 2.00 inches = 2.00/12 feet = 0.1667 feet step 2: calculate the volume in cubic feet volume = area × height volume = 1.00 square mile × 0.1667 feet = 27,878,400 cubic feet step 3: convert cubic feet to gallons 1 cubic foot = 7.48052 gallons volume in gallons = 27,878,400 cubic feet × 7.48052 gallons/cubic foot ≈ 208,551,817 gallons so, approximately 208,551,817 gallons of water would correspond to 2.00 inches of rainfall across 1.00 square mile. question 5: converting flow rate to convert a flow rate of 1.83 ml/hour to units of l/second: step 1: convert ml to l 1 ml = 0.001 l 1.83 ml/hour = 0.00183 l/hour step 2: convert hours to seconds 1 hour = 3600 seconds 0.00183 l/hour = 0.00183 l/3600 seconds ≈ 5.0833 x 10^-7 l/second so, the flow rate of 1.83 ml/hour is approximately 5.0833 x 10^-7 l/second. question 6: fraction of molecules escaping to calculate the fraction of water molecules at the surface of the ocean with sufficient kinetic energy to escape into the atmosphere, we can use the boltzmann distribution of kinetic energy: the boltzmann distribution equation is:

f (e) = e^{(} - e / kt)

where:

f (e)

is the fraction of molecules with energy

e

or higher.

e

is the energy required for escaping (44.0 kj/mol in this case).

k

is the boltzmann constant (8.617333262145 x 10^-5 ev/k or 1.380649 x 10^-23 j/k).

t

is the temperature in kelvin. for

t = 30° c = 303.15 k

:

f (e) = e^{(} - 44000/ (1.380649 x 1 0^{-} 23 * 303.15))

calculate the value of

f (e)

using the above equation. repeat the calculation for

t = 20° c = 293.15 k

. draw two boltzmann distribution curves, one for each temperature, and shade in the areas that represent the fraction of molecules with sufficient energy to escape at each temperature. unfortunately, i can't draw diagrams in this text-based format, but you can plot the curves using graphing software or tools like microsoft excel. remember that the fraction of molecules with sufficient energy to escape will be higher at the higher temperature (30°c) compared to the lower temperature (20°c) due to the exponential relationship in the boltzmann distribution equation.

coulombic potential energies for charge pairs: a. +2 ion (q1 = +2e) separated by 10 nm (d1 = 10 nm) from a +2 charge (q2 = +2e): coulombic potential energy: pe1 = k(q1q2/d1) pe1 = (9.0 x 10^9 nm²/c²)((2e)(2e)/(10 nm)) note: e represents the elementary charge, which is 1.602 x 10^-19 c pe1 = (9.0 x 10^9 nm²/c²)(4e²/(10 x 10^-9 m)) pe1 = (9.0 x 10^9 nm²/c²)(4(1.602 x 10^-19 c)²/(10 x 10^-9 m)) pe1 = (9.0 x 10^9 nm²/c²)(4(2.567 x 10^-38 c²)/(10 x 10^-9 m)) pe1 = (9.0 x 10^9 nm²/c²)(1.027 x 10^-37 c²/10 x 10^-9 m) pe1 = (9.0 x 10^9 nm²/c²)(1.027 x 10^-28 c²/m) pe1 = 9.243 x 10^-19 j +3 ion (q1 = +3e) separated by 5 nm (d2 = 5 nm) from a +1 ion (q2 = +1e): coulombic potential energy: pe2 = k(q1q2/d2) pe2 = (9.0 x 10^9 nm²/c²)((3e)(1e)/(5 nm)) pe2 = (9.0 x 10^9 nm²/c²)(3e²/(5 x 10^-9 m)) pe2 = (9.0 x 10^9 nm²/c²)(3(1.602 x 10^-19 c)²/(5 x 10^-9 m)) pe2 = (9.0 x 10^9 nm²/c²)(3(2.567 x 10^-38 c²)/(5 x 10^-9 m)) pe2 = (9.0 x 10^9 nm²/c²)(7.701 x 10^-38 c²/5 x 10^-9 m) pe2 = (9.0 x 10^9 nm²/c²)(1.5402 x 10^-28 c²/m) pe2 = 1.386 x 10^-18 j the magnitude of coulombic potential energy for the +3 ion is approximately 1.5 times greater than that for the +2 ion. b. a-3 ion (q1 = -3e) separated by 6 nm (d1 = 6 nm) from a +1 ion (q2 = +1e): coulombic potential energy: pe1 = k(q1q2/d1) pe1 = (9.0 x 10^9 nm²/c²)((-3e)(1e)/(6 nm)) pe1 = (9.0 x 10^9 nm²/c²)(-3e²/(6 x 10^-9 m)) pe1 = (9.0 x 10^9 nm²/c²)(-3(1.602 x 10^-19 c)²/(6 x 10^-9 m)) pe1 = (9.0 x 10^9 nm²/c²)(-3(2.567 x 10^-38 c²)/(6 x 10^-9 m)) pe1 = (9.0 x 10^9 nm²/c²)(-7.701 x 10^-38 c²/6 x 10^-9 m) pe1 = (9.0 x 10^9 nm²/c²)(-1.2835 x 10^-29 c²/m) pe1 = -1.155 x 10^-19 j -2 ion (q1 = -2e) separated by 5 nm (d2 = 5 nm) from a +2 ion (q2 = +2e): coulombic potential energy: pe2 = k(q1q2/d2) pe2 = (9.0 x 10^9 nm²/c²)((-2e)(2e)/(5 nm)) pe2 = (9.0 x 10^9 nm²/c²)(-2e²/(5 xapologies, but it seems there might be some missing information or incomplete sentences in the question. the given statements are not clear and lack the necessary details to perform the calculations. can you please provide the complete and accurate information for each question?

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