Solve the following problems. Enter numerical result and units in space provided

Question

Solve the following problems. Enter numerical result and units in space provided on this cover sheet. Express results to three or more significant figures. Attach supporting hand-written work to this cover sheet. Prob. 1 If the diameter of one hair follicle is 50mu. How many hair follicles would it take to 'cover' the surface of the Uranus having a radius of 25560 km? Assume 'closest packing' of hair follicles (akin to packing of a rack of billiard balls). Express the result in Avogadro units where 6.022E23 follicles = 1.000 Avogadro unit of follicles. Quantity = Prob. 2 A component is composed of 0.05140 moles of nickel and 0.01990 moles of silver. Determine (a) total mass of the component (b) weight % of nickel Prob. 3 A relic having a mass of 320 g is known to contain 89.500 wt. % nickel and 10.500 wt % platinum. Determine the atomic percentage of nickel in the relic. Prob. 4 An energy input of 33.44kJ is required to raise the temperature of 100 grams of water from room temperature (20degreeC) to boiling temperature (100degreeC). How many photons of energy are needed to bring 170 grams of room temperature water to boiling if 276 mili-meter electro-magnetic radiation is used to heat the water? Prob. 5 The repulsive force between two ions. A^1+ and B1^- at equilibrium is 1.100E-8 N. Calculate the following (Assume n =7 in Coulomb's law): DON'T FORGET UNITS!!! (a) interionic distance: (b) iconic radius for B^1- it that for A^1+ is 0.116 nm: (c) attractive force between the two ions at this position: (d) value of the constant b: (e) minimum potential energy of the bond: Prob. 6 For each of the following compounds, use Pauling's equation to estimate the percentage ionic character

Explanation / Answer

Problem 2) Weight of Ni= no of mole of Ni * Atomic weight of Ni = 0.0514 mole* 58.6934 gr/mole = 3.0168 gr

Weight of Ag= no of mole of Ag * Atomic weight of Ag = 0.0199 mole* 107.8682 gr/mole = 2.1465 gr

a) Total weight = 3.0168 gr+2.1465 gr = 5.1633 gr

b) Weight % of Ni = (wt of Ni/ Total wt) * 100 =(3.0168/5.1633) * 100 = 58.42 %

Problem 3) 100 gr of compound contains 89.5 gr of Ni, So 320gr contains===> (320 * 89.5)/100 = 286.4 gr

286.4 gr of Ni = 286.4 gr/58.693 gr/mole = 4.79 mole of Ni.

100 gr of compound contains 10.5 gr of Ag, So 320gr contains===> (320 * 10.5)/100 = 33.6 gr

33.6 gr of Ag = 33.6 gr/107.8682 gr/mole = 0.311 mole of Ag.

Total mole = 4.79 + 0.311 =5.10 mole.

Atomic % or Mole % of Ni = (4.79/5.10) *100= 93.9 %

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