How many significant figures are in the measurement 40,500 mg? a.|two|c.|four| b.|three|d.|five| 5. Express the sum of 7.68 m and 5.0 m using the correct number of significant digits. a.|12.68 m|c.|13 m| b.|12.7 m|d.|10 m| 6. What is the measurement 111.009 mm rounded off to four significant digits?
Volumetric Pipette 10 ml Unknown liquid was measured in a 150 ml beaker. Using a volumetric pipette to measure 10 ml, 20 ml, 30 ml respectively. The average density: =0.750+ 0.727+ 0.720 = 2.197 = 2.197/3 = 0.73 g/ml Method 3. Burette Unknown liquid was measured into a 150 ml beaker and using a burette to measure 13 ml, 26 ml, and 39 ml of the liquid and recorded masses. The average density: =0.82+ 0.77+ 0.77 = 2.36 = 2.36/3= 0.79 g/ml Note that the burette was the most accurate in determining accuracy of measured amounts.
1.614 M; c. 0.538 M; d. 2.152 M; 3. 100 times; 4a. 2.6, acidic; b. 2 x 10-10 M, basic; c. 5.01 x 10-11 M, basic; d. 9.61, basic; e. 3.2 x 10-9 M, acidic; 5a. Na2SO4 → 2 Na+ + SO42-; b.
|50.08[pic] 10[pic] km |d. |5.008 [pic] 10[pic] km | ____ 3. What is the result of multiplying 2.5 [pic] 10[pic] by 3.5 [pic] 10[pic]? |a. |8.75 [pic] 10[pic] |c.
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:
Reflection questions about the Drake equation: Equation | Minimum values | Maximum values | R= | 1 | 7 | Fp= | 0.4 | 0.6 | Ne= | 2 | 2.5 | F1= | 0.5 | 1 | Fi= | 0.001 | 1 | Fc= | 0.5 | 0.8 | L= | 10,000 | 200,000 | N= | 2 | 1,680,000 | What value did you get for the number of civilizations? After calculating the maximum and minimum values of the equation through researching them individually, the minimum value of the Drake equation, N = R x fp x ne x f1 x fi x fc x L, was N = 2. The values were as such: * R = 1 * fp = 40 % (0.4) * ne = 2 * f1 = 50% (0.5) * fi = 0.001 * fc = 50% (0.5) * L = 10,000 (1 x 0.4 x 2 x 0.5 x 0.001 x 0.5 x 10000). These figures lead the end of the equation at the number of communicative civilizations at N = 2; meaning there is a minimum of 2 expected communicative civilizations in the galaxy. The maximum value that was proposed for the Drake equation, N = R x fp x ne x f1 x fi x fc x L, was N = 1,680,000.
Riboflavin 3.4 mg 200% 9. Niacin 20 mg 100% 10. B6 6 mg 300% 11. Folic acid 400 mcg 100% 12. B12 25 mcg 417% 13.
During the 30 min waiting period, the TLC and vacuum filtration apparatus was setup. 5. Even though the lab procedure indicated that less than 25 mL of HCL was needed, we ended up using about 27 mL of HCL. 6. The litmus paper turned a faint red after 27.5 mL of HCL was added.
| Lucy and Jean have 240 stickers altogether. If Lucy gives 10 of her stickers to Jean, she will have three times as many stickers as Jean. How many stickers does Lucy have? | 6.
Your answer : b. The hemoglobin levels for these two individuals will be the same. Stop & Think Questions: Why is the average hematocrit higher in males than in females? You correctly answered: c. Higher testosterone levels in males promotes more RBC production. Experiment Data: Blood sample 1 2 3 4 5 gm Hb per 100 ml blood 16 14 8 20 22 Hematocrit (PCV) 48 44 40 60 60 Ratio of PCV to Hb 3:1 3.14:1 5:1 3:1 2.73:1 11/05/14 page 2 Post-lab Quiz Results You scored 100% by answering 3 out of 3 questions correctly.