20. mol H2 reacts with 8.0 mol O2 to produce H2O. Determine the number of grams reactant in excess and number of grams H2O produced. Identify the limiting reactant. 8.1 g H2 , 2.9 x 102 g H2O 17. How many litres of O2 gas are required to produce 100. g Al2O3?
5) Information about Clearwater Company's direct materials cost follows: Standard price per materials ounce $ 100 Actual quantity used 8,700 grams Standard quantity allowed for production 9,100 grams Price variance $ 76,125 F ________________________________________ Required: What was the actual purchase price per gram? (Round your answer to 2 decimal places. Omit the "$" sign in your response.) Actual purchase price $ 91.25 Total grade: 0.0×1/1 = 0% Feedback: Actual Costs = AP × 8,700 Actual Inputs at Standard Price = $100 × 8,700 =$870,000 Price Variance = $76,125 F 8,700 × AP = $870,000 – $76,125 AP = $91.25 ________________________________________ Question 3: Score
Over 16,000,000 3. What would be the dotted decimal equivalent o the slash notation of /30? 30/8= 3r6 255.255.255.252 4. What would be the dotted decimal equivalent o the slash notation of /8? 8/8= 1 255.0.0.0 5.
Practice Question for Exam 2 Chem 103 |A chemical reaction requires 30.77 kJ. How many kilocalories does this correspond to? | |A) 7,354 kcal B) 7.354 kcal C) 128.7 kcal D) 0.1287 kcal E) 30.77 kcal | |776 J is the same quantity of energy as | |A) 7.76 × 105 kJ. B) 1.85 × 102 kcal. C) 0.185 kcal.
Part I: Density of Unknown Liquid | | Trial 1 | Trial 2 | Trial 3 | Mass of Empty 10 mL graduated cylinder (grams) | 25.5g | 25g | 25g | Volume of liquid (milliliters) | 8.6mL | 8.7mL | 8.4mL | Mass of graduated cylinder and liquid (grams) | 36g | 36g | 35.5g | Part II: Density of Irregular-Shaped Solid | Mass of solid (grams) | 38.74g | 39.002g | 42.489g | Volume of water (milliliters) | 50mL | 49mL | 51mL | Volume of water and solid (milliliters) | 54mL | 53mL | 56mL | Part III: Density of Regular-Shaped Solid | Mass of solid (grams) | 26g | 27g | 26g | Length of solid (centimeters) | 5.2cm | 5cm | 4.5cm | Width of solid (centimeters) | 3cm | 4cm | 3.5cm | Height of solid (centimeters) | 2.5cm | 3cm | 2cm | Calculations Show all of your work for each of the following calculations and be careful to follow significant figure rules in each calculation. Part I: Density of Unknown Liquid 1. Calculate the mass of the liquid for each trial. (Subtract the mass of the empty graduated cylinder from the mass of the graduated cylinder with liquid.) * Trial 1 36-25.5=10g * Trial 2 36-25=11g * Trial 3 35.5-25=10g 2.
Why or why not? | Not unusual, because it is within 2 standard deviations of the mean | 8d. Fifty adults are randomly selected. What is the likelihood that the mean of their body temperatures is 97.98°F or lower? | 0.62 percent | 9a.
Laboratory Techniques and Measurements – Lab Report Assistant Exercise 1: Length, Temperature, and Mass Data Table 1. Length Measurements Object Length (cm) Length (mm) Length (m) CD or DVD 12 120 .12 Key 3.9 39 .039 Spoon 157. 157 .157 Fork 18 180 .18 Data Table 2. Temperature Measurements Water Temperature (°C) Temperature (°F) Temperature (K) Hot from tap 50 122 323.15 Boiling 100 212 373.15 Boiling for 5 minutes 123 253 396.15 Cold from tap 20 66 293.15 Ice water – 1 minute 12 46 283.15 Ice water – 5 minutes 4 39 278.15 Data Table 3. Mass Measurements Object Estimated Mass (g) Actual Mass (g) Actual mass (kg) Pen or pencil 4 4.7 .0045 3 Pennies 3 7.5 .0075 1 Quarter 5 6.3 (1964 silver) .0063 2 Quarters, 3 Dimes 13 19.2 .0192
1a) Calculate molar solubility of SrF2 in: 0.010 M Sr(NO3)2 SrF2 <=> Sr+2 & 2 F- Ksp = [Sr+2] [F-]^2 4.3x10^-9 = [0.010 M] [F-]^2 [F-]^2 = (4.3x10^-9) / [0.010 M] [F-]^2 = 4.3x10^-7 [F-] = 6.56 X 10^-4 Molar by the equation: 1 mole of SrF2 <=> Sr+2 & 2 moles of F- the molar solubility of SrF2 that releases [F-] is half as much: 3.279 X 10^-4 Molar SrF2 yoiur answer, rounded to 2 sig figs would be 3.3 X 10^-4 Molar SrF2 ======================================… 1b) Calculate molar solubility of SrF2(Ksp=4.3x10^-9) in 0.010 M NaF Ksp = [Sr+2] [F-]^2 4.3x10^-9 = [Sr+2] [0.010]^2 [Sr+2] = 4.3x10^-9 / [0.010]^2 [Sr+2] = 4.3x10^-9 / 1 X 10^-4 [Sr+2] = 4.3x10^-5 Molar by the equation:
Math 015 FINAL REVIEW Name____________________ 1. Write in expanded form 7,982 a. 70,000 + 900 + 80 + 2 b. 7,000 + 900 + 80 + 2 c. 7,000 + 9000 + 80 + 2 d. 7,000 + 900 + 80 + 20 2. Round 689,652 to the nearest thousands a.
Results - After transferring the distillate to a reaction group the halide form the top layer. The final weight of the halide was 0.17 g. % yield = (experiment yield / Theoretical yield) * 100 Calculating theoretical yield: 1 ml butanol * (0.81 g butanol / 1 ml butanol) * (1 Mole butanol / 74.12 g butanol) * (137.03 bromobutane /1 mole bromobutane) = 1.497 g of 1-bromobutane. 1.33 g of NaBr * (1 mole NaBr / 102.91 g NaBr) * (1 mole bromobutane / 1 mole NaBr) * (137.03 g bromobutane / 1 moe bromobutane) = 1.77 g of 1-bromobutane The result shows that the 1-butanol is the limiting reagent so the theoretical yield is 1.497 g of 1-bromobutane. % yield = (0.17 / 1.497)* 100 =