CHEM 120 Unit 4 Midterm | Questions and Answers | (Latest Update) | 100% CORRECT SOLUTIONS.

CHEM 120 Unit 4 Midterm 1. The correct isotope symbol for the isotope containing 1 proton and two neutrons – He 2. Which are the elements with the following electronic configuration: 1s2 2s 2 2p6 3s1 – Sodium 1s2 2s2 2p5- Fluorine 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p2 – Geremanium (Ge) 3. Provide 4 examples of substances with ionic bonds LiF - Lithium Fluoride, LiCl - Lithium Chloride, LiBr – Lithium Bromide, LiI - Lithium Iodide, Nacl,Calcium Nitrate 5. Name the following: N2S3 - dinitrogen trisulfide NaCl – Sodium Choloride P4O10- Phosphorus pentoxide NaNO3- Sodium nitrate Cl2O3- Dichlorine trioxide 6. A gas is enclosed in a 10.0 L tank at 350 mm Hg pressure. If the gas is pumped into a 23.00 L vessel, what is the new pressure 10.0 L×350 mmHg 23.00 L = 10.0 × 350mmHg 23.00 = 3500 mmHg 23 152.173 mmHG 7. The molarity of a solution that contains 12.5 g of KClO3 in 3.2 liter of solution is Molarity = moles of solute volume of solution ; K- 39.10 + Cl-35.45 + O- 16(3)-48; KClO3= 122.55 g/mol M= 12.5 g×1 mol 122.5 g = 12.5 mol 122.5 = 0.102 mol 0.102mol 3.2 L = 0.032 M 8. What other compound could also be added in order to make a buffered solution with each of the following: H3PO4 - H2PO3 CH3COOH - CH2CO - H2CO3 - HCO - 9. Balance: C2H4 + O2 → CO2 + H2O First take a look at the equation and see how many elements there are, in this case we have 3 different elements; Carbon, Hydrogen and Oxygen. To balance the equation start off by balancing the elements in the complex molecules first and then the single element molecules after. We'll start with balancing the Carbons. On the left we have 2, but on the right only 1, so we multiply the CO2 on the right by 2. Next we look at the Hydrogen. We have 4 on the left, but only 2 on the right. To balance them, we multiple the H2O by 2. Now we have to take a look at the Oxygen. From balancing the Hydrogen and Carbon, we have changed the initial equation. No looking at the Oxygen on the right-hand side, we have 4 from the CO2 and 2 from the H2O for a total of 6. In order to have 6 Oxygen on the left as well we have to multiply O2 by 3. Now we have the balanced chemical equation of: C2H4 + 3O2 --> 2CO2 + 2H2O 10. For the balanced chemical reaction: 4Al(s) + 3O2(g) → 2Al2O3(s) a. Determine the molar mass of Al2O3. Aluminum oxide has a molar mass of 101. g/mol. b. Determine how many grams of Al2O3 you would require to produce 7 moles of Al(s) c. 7 molAl (s )× 2 molAl2 O3 × 101.96 g = 713.72 g Al O 1 molAl ( s) ×1 molAl2O3 11. List the number of protons, neutrons, and electrons for 79Br- Protons 35 , 44 Neutrons, Electrons 35 12. An isotope has 21 protons and 19 neutrons. What would be the mass number of this isotope 40 13. What are the correct formulas for the compounds formed between calcium ion and the sulfate ion : CaSO4 calcium ion and nitrate ion: Ca(NO3)2 calcium ion and sulfite ion : CaSO4 calcium ion and hydroxide ion : Ca(OH)2 14. Compute the pH of the following solutions with the [H3O+] = 1.2 x10-3 - [H3O] = 1.2 x10-3 pH= 3 [H3O+] = 1x10-6 pH= 6 15. At what temperature will 0.654 moles of neon gas occupy 12.30 liters at 1.95 atmospheres? • P (pressure), which is 1.95 atm. • V (volume), which is 12.30 L. • n (number of moles), which is 0.654 moles of neon gas. We have to find T. To do this, we'll need to use the Ideal Gas Law, which is: PV=nRT Rearranging this equation to get temperature on one side, we get: T=PVnR Pressure is in atm and volume is in L. This tells us that we'll need to use the value of 0.08206 L atm/K mol for R, the ideal gas constant. T=pVnR T=1.95atm×12.30L0.654mol×0.08206LatmK−1mol T=447K We can then convert to celsius, which equals to around 174∘C. 16. Determine the volume of occupied by 2.34 grams of carbon dioxide gas at STP. In order to solve this problem we would use the Ideal Gas Law formula 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 2.34gCO2 must be converted to moles 2.34gCO2x1mol44gCO2=0.053mols P=1atm V=???L n=0.053mols R=0.0821atmLmolK T=273K PV=nRT becomes V=nRTP V=0.053mols(0.0821atmLmolK)(273K)1atm V=1.18L 17. Calculate how many grams are in 0.700 moles of H2O2 23.811 grams 18. Convert 2.50 moles of KClO3 to grams 0.. Show Less