Bonus Assignment Questions

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Carleton University *

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1001

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Chemistry

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Dec 6, 2023

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docx

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Uploaded by ConstableSummer12429

Bonus Assignment This is a mock assignment. So long as you attempt every question, you will receive a bonus 3% to your final grade. This is not timed. You can take as much time as you need, so long as you submit your assignment by 11:59PM Friday December 2 nd This is open book, not copy paste. You can consult online resources (your notes, tutorials, textbook, etc…) but do not plagiarize off the internet. Yes, I know how google works, and yes, I will find out. Triple check your calculations. Calculators are machines that will do what you tell them, so make sure you put the right numbers in so you get the right numbers out. Take your time, and show your work. This is supposed to help prepare you for your final exam. I can’t give feedback if you only provide the answer. Write clearly, we mark what we read. You can’t mark what you can’t see. Good luck, you got this. Total (/100) 1 (/10) 2 (/10) 3 (/10) 4 (/10) 5 (/10) 6 (/10) 7 (/10) 8 (/10) 9 (/10) 10 (/10)

1. 2.0 mol of H 2(g) is reacted with 250g of I 2(g) to form HI (g) in a 2.00L vessel. If the equilibrium constant (K eq ) is 3.50*10 3 , what is the mole fraction of HI (g) in the vessel at equilibrium. 4 x 2 ( L − x ) 2 = 3.5 × 10 3 2 x L − x = √ 3.5 × 10 3 = 59.2 2 x = 59.2 − 59.2 x 2 x + 59.2 x = 59.2 x = 59.2 2 + 59.2 x = 0.967 Molarity of HI at equilibrium = 0.967 × 2 = 1.934 M No. of moles of HI = 2 × 1.934 moles = 3.868 moles Mole fraction of HI = 3.868 3.868 + 0.066 + 0.066 = 0.967 0.066 is the moles of H2 and I2 at equilibrium L-n=1-0.967 n=0.033M Therefore, number of moles =0.033mol/L x 2L=0.066mole

2. 1.172kg of an unknown gas comprised of only carbon and hydrogen is held in a 50.0L vessel at 25.0 o C. If it exerts a pressure of 10 bar, what is the molecular weight of this gas? Based on the molecular weight, what is the identity of this gas? PV=nRT n=Pv/RT = 9.86atmx 50.0L/0.082Latm mol^-1 x 298K =20.18mol Molecular weight of hydrocarbon =58.19/mol 3. Using the Rydberg equation, calculate the binding energy of a 1s electron in hydrogen. E= -1.09*10^7 x 6.6*10^34 x 3*10^8 (1/1) E= 1.0967 x 3*10^-19 J E= 3.30*10^-19 J Therefore binding energy is 3.30*10^-19 J for 1s electron in hydrogen

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4. Sodium hydroxide reacts with phosphoric acid to form sodium phosphate and water: 3 NaO H ( aq ) + H 3 PO 4 ( aq ) → N a 3 PO 4 ( aq ) + 3 H 2 O ( l ) If a Carleton University student adds 10.00 grams of NaOH to react with 10.00 grams of H 3 PO 4 … a) What is the limiting reactant? NaOH b) How many grams of Na 3 PO 4 are formed (assuming the reaction goes fully to completion)? 13.67g c) How many grams of the excess reactant remains when the reaction is complete (assuming the reaction goes fully to completion)? 1.83g 5. The human eye can detect as little as 2.5*10 -16 joules of light. How many photons of green (510nm) light can your eye detect? How many photons of red (620nm) light can your eye detect? Wavelength of green light =510nm = 510 × 10 − 9 m Energy of light ¿ hc λ = 6.636 × 10 − 34 J ∙s× 3 × 10 8 m / s ❑ 510 × 10 − 9 m = 3.90 × 10 − 19 J

No. of green photon ¿ total energy energy of one photon = 3.90 × 10 − 19 J 2.5 × 10 − 16 = 1.56 × 10 − 3 Therefore, no. of green photon is 0 because it can’t be a fraction Energy of light ¿ hc λ = 6.636 × 10 − 34 J ∙s× 3 × 10 8 m / s ❑ 620 × 10 − 9 m = 3.21 × 10 − 19 J No. of red light photon ¿ 3.21 × 10 − 19 J 2.5 × 10 − 16 J = 1.28 × 10 − 3 Therefore, no. of red light photon is zero because it cannot be a fraction 6. An unknown metal sample is interrogated using a beam of X-rays to liberate electrons which are then caught by a detector that determines their kinetic energy. A beam of X-rays with wavelength 9.54nm irradiate the surface of this metal, and the detector observes electrons with kinetic energies equal to 7.3716*10 -18 J. What is the binding energy of this unknown metal sample. For the x ray, Wavelength, λ = 9.54 nm = 9.54 × 10 − 9 m Energy, E = hc λ [h= Planck’s constant = 6.626 × 10 − 34 J ∙s ] [c= velocity of light = 3 × 10 10 cm / s ]

¿ 3 × 10 8 m s ¿ ( 6.626 × 10 − 34 J ∙s ) × ( 3 × 10 8 m / s ) ( 9.54 × 10 − 9 m ) ¿ 2.084 × 10 − 17 J bindingenergy of the metalsample = ( energyof xray ) − ( kinetic energyof electron ) = ( 2.084 × 10 − 17 J ) − ( 7.3716 Therefore, the binding energy of the unknown metal sample is 1.3465 × 10 − 17 J 7. What is the De Broglie wavelength of an electron going 99% the speed of light? 0.24nm

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8. Determine the volume occupied by 2.34g of CO 2 under STP conditions. PV=nRT P= pressure in atm V= volume in L N= moles R= ideal gas law constant T= temp in K STP is standard temperature and pressure which has values of 1atm and 273K 2.34 CO 2 must be converted to moles 2.34 gCO 2 x 1 mol 44 gCO 2 = 0.053 mols P= 1atm V=? L n= 0.0821 atmL molK T= 273K PV= nRT becomes V= nRT P v = 0.053 mols ( 0.0821 atmL mol K ) ( 273 K ) 1 atm V= 1.18L

Alternative solution: 2.34 ∙ 1 mol 44 gCO 2 = 0.053 mol 22.4 L mol ∙ 0,053 mol = 1.18 L 9. Chlorine has two stable isotopes. Based on the following information, what is the molecular weight of Cl? Isotope mass 34.969 amu 36.966 amu Abundance 76% 24% Cl molecular weight = 70.906 g/mol 10. Your local radio station Jump! 106.9FM operates at a radio frequency of 106.9 MHz. Your cars radio receives a signal at a rate of 4.807 *10 -21 J/s. Assuming your favourite song lasts 3.50 minutes, how many photons does your car receive from start to finish?

Energy associated with waves; E w = hc λ =6.626 × 10 − 34 × 106.9 × 10 6 = 7.08 × 10 − 26 J ∵ MHz = 10 6 Hz 1 Hz = 1 s − 1 Total energy; E t = rate×time =4.807 × 10 − 21 × 3.50 × 60 ( ∵ 1 min = 60 s ¿ No. of photons received = E t E w ¿ 1.01 × 10 − 18 J 7.08 × 10 − 26 J N = 1.425 × 10 7 Therefore, the car receives 1.425 × 10 7 photons

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