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BIO560 LAB 1 REPORT RECENTLY UPDATED (PASS ASSURANCE)

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BIO560 LAB 1 REPORT
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BIO560 LAB 1 REPORT

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UNIVERSITI TEKNOLOGI
MARA FACULTY OF
APPLIED SCIENCES


BACHELOR OF SCIENCE (HONOURS)
BIOLOGY




BIO560 LAB 1
REPORT


FUNDAMENTAL PHYSIOLOGICAL
PRINCIPLES



LECTURER: NOR LAILATUL WAHIDAH
MUSA
NAME: MUHAMAD QAMARUL AMSYAR BIN MOHD HAZANI
SID: 2021843536

GROUP: AS2013A

DATE: 28th OCTOBER 2022

GROUP MEMBERS:
1. MUHAMAD DINIE ASHRAF BIN MOHAMMAD DASUKI
2. AHMAD ALIF HAIKAL BIN MAMAT
3. MUHAMMAD DANIEL HAKIM BIN ROSLI.

,POST LAB QUESTION

A. Units of Measurement

a. Provide the correct conversion units for the following
measurements: 100 Å thickness of cell membrane = 10 nm
1A = 0.1nm
100A x 1A = 100A
100A x 0.1 = 10nm

8 Å diameter membrane pore = 800 pm
1pm = 0.01A
.01 = 800pm

15 g hemoglobin per 100 ml blood =
150mg/ml 15g = 15 x 1000 = 15000mg
15000 mg / 100 ml = 150 mg/ml

100 ml plasma =
0.1L 1000ml = 1L
100/ 1000 = 0.1L

25°C room temperature = 77 °F
0° C = 32°F
Formula = ( °C X 9/5) + 32
= 77°F

98.6°F body temperature = 37 °C
0° C = 32°F
Formula = (32°F - 32) x 5/9 = 0°C
= (98.6°F - 32) x 5/9 = 37°C


b. A powerlifter lifts 500 lb 6 ft off the ground. How much work has he performed?

3000 ft-lb work = 414.9 kg-m work

3000 lb/ft = 1 kg-m / 7.23 lb/ft
= 414.9 kg-m

,How many calories of energy did he used to produce this work? 973 cal
1 kg-m = 2.34385 cal
414.94 x 2.34385 = 973 cal

If he performs this feat 11 times in 1 minute, what is his power output? 746 W
1 Horsepower (HP) = 746 W

, B. Concentration of Solutions




1. How many grams of glucose would you need to make 500 ml of an 8 % solution?

40 g

percentage = (gram solute / volume solution)
x 100 glucose (g) = () x 500
= 40g

2. If 6 g of NaCl is dissolved in 1 L of solution, what percent concentration is prepared?

0.6 %

percentage concentration = 6g / 1000 ml x 100
= 0.6 %

3. How many grams of KCl would you need to make 250 ml of a 0.5 M solution?

9.31 g

M = n/V
0.5 = n/0.25
n = 0.125 mol

Molar mass of KCL= 39 g/mol (K) + 35.5 g/mol (Cl) = 74.5 g/mol.
= (39+35.5) 0.125 mol
= 9.3125 g
= 9.31 g
+

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BIO560 LAB 1 REPORT

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