Exercise 10: Acid-Base Balance: Activity 3: Renal Responses to Respiratory Acidosis and Respiratory Alkalosis Lab Report Pre-lab Quiz Results You have not completed the Pre-lab Quiz. 07/18/13 page 1 Experiment Results Predict Question: Predict Question 1: What effect do you think lowering the PCO2 will have on [H+ ] and [HCO3- ] in the urine? Your answer : d. [H+ ] will decrease and [HCO3- ] will increase. Predict Question 2: What effect do you think raising the PCO2 will have on [H+ ] and [HCO3- ] in the urine? Your answer : c. [H+ ] will increase and [HCO3- ] will decrease.
2) Percent recovery for isolation of benzoic acid % Recovery = mass of recovered material _________________________________ x100% mass of starting material = (0.43/1.01) x100% = 42.57% That concludes that the percent recovery is 42,57%. 3) Percent recovery for isolation of hydroquinone dimethyl ether % Recovery = mass of recovered material _________________________________ x100% mass of starting material = (0.16/1.01) x100% = 15.84% That concludes that the percent recovery is 15.84%. Table 2: : Experimental IR peaks compared to literature IR peaks for Benzoic acid Functional groups | Experimental peak (cm-1) | Literature peak (cm-1) | O-H | 3407-2563 | 3400-2564 | C=O | 1689 | 1689 | C-H |
Gas Law Problems - Graham's Law Return to Graham's Law Return to KMT & Gas Laws Menu Problem #1: If equal amounts of helium and argon are placed in a porous container and allowed to escape, which gas will escape faster and how much faster? Solution: Set rate1 = He = x Set rate2 = Ar = 1 The molecular weight of He = 4.00 The molecular weight of Ar = 39.95 Graham's Law is: r1 over r2 = √MM2 over √MM1 Substituting, we have: x / 1 = √(39.95 / 4.00) x = 3.16 times as fast. Video: Solution to a Graham's Law Problem Problem #2: What is the molecular weight of a gas which diffuses 1/50 as fast as hydrogen? Solution: Set rate1 = other gas = 1 Set rate2 = H2 = 50 The molecular weight of H2 = 2.02 The molecular weight of the other gas = x. By Graham's Law (see the answer to question #1), we have: 1 / 50 = √(2.02 / x) x = 5050 g/mol Video: Solution to a Graham's Law Problem Problem #3: Two porous containers are filled with hydrogen and neon respectively.
(c) If you wanted 95 percent of the kernels to pop, what time would you allow? (d) If you wanted 99 percent to pop? A) 2 minutes = 120 seconds z(120) = 120-150)/25= -30/25 = -6/5= -1.2 P(z>-1.2) = 0.1151 B) 3 minutes = 180 seconds z(180) = (180-150)/25=30/25 = 6/5 = 1.2 P(z>1.2) = 0.1151 C) If you want 95% of the kernels to pop, what time would you allow? (.95) = 1.645 x-150 = 25*1.645 x = 41.12 = 150= 191.12 seconds x = 3 minutes and 11.12 seconds D) If you want 99% to pop… (.99) = 2.326 2.326 = (x-150)/25 x = 25* 2.326 + 150 x= 208.16 seconds x = 3 minutes and 28.16 seconds 8.46 A random sample of 10 miniature Tootsie Rolls was taken from a bag. Each piece was weighed on a very accurate scale.
+ 102 C. + 350 D. + 394 (1) 7. What is the energy change (in kJ) when the temperature of 20 g of water increases by 10°C? A. 20 × 10 × 4.18 B. 20 × 283 × 4.18 C. D.
State the place value of the underlined digit in 6 953 742. A Hundreds C Ten thousands B Thousands D Hundred thousands ( 3. Round off 5 987 341 to the nearest hundred thousand. A 5.8 million C 6.0 million ( B 5.9 million D 6.1 million 4. Which of the following numbers, when rounded off to the nearest thousand, becomes 7 541 000?
For example, when the amount of gas was doubled from 5.0 mL to 10.0 mL, the pressure is halved and the reading should have been 96.32 kPa as opposed to 104.54 kPa. This experiment correlated with the material present in our textbook in
| 6.99V | 0.14A | 7. | 8.00V | 0.16A | 8. | 9.03V | 0.18A | 9. | 9.98V | 0.20A | 10. | 18.01V | 0.36A | 50-ohm Resistor | V drop | I through | 1.
The angle of the drop height was at 90 degrees and the length of the string was at 300 mm. Size of Mass ( grams) | Trial 1 (seconds) | Trial 2 (seconds) | Trial 3 (seconds) | Average (seconds) | Period (seconds) | 10 grams | 25.00 seconds | 24.90 seconds | 25.21 seconds | 25.04 seconds | 1.252 seconds | 100 grams | 25.86 seconds | 25.93 seconds | 25.81 seconds | 25.87 seconds | 1.2935 seconds | 500 grams | 26.53 seconds | 26.30 seconds | 26.44 seconds | 26.42 seconds | 1.321 seconds | 1,000 grams | 25.82 seconds | 26.46 seconds | 25.91 seconds | 26.06 seconds | 1.303 seconds | 4.) Angle or Drop Height: Hypothesis: I predict that the larger the angle of drop, the longer period of the pendulum. The independent variable is the drop height. The mass of the pendulum was 100 grams and the string measured 300 mm.
In each trial, the initial reading, final reading and the volume of HCl used was recorded down as quantitative results. The average volume of hydrochloric acid was found to be 12.03mL. The amount of sodium carbonate in the 10.00ml of solution was found to b 0.05 mol. The amount of hydrogen chloride that was dissolved in the average volume of acid is 0.365g. Through these calculations, the concentration of hydrochloric acid was found to be 8.3 mol•L-1.