Question

name cristabel cortez
date
desk
laboratory instructor dr. car
report sheet experiment 1
basic laboratory techniques
a. the meterstick
length of this lab book 10.8 in. 7.5 cm 75 mm.075 m
width of this lab book 8.7 in. 22.1 cm 221 mm.221 m
using an equation (including units), show that the above measurements are equivalent.
1 in = 2.54 cm
1 cm = 10 mm
1 m = 100 cm
area of this lab book (show calculations) 160 cm²
b. the graduated cylinder
volume of water in graduated cylinder 100 ml
volume of water contained in largest test tube 30 ml
c. the thermometer and its calibration
observed temperature of water - and - ice mixture 5.4 °c
temperature of boiling water 99 °c
observed atmospheric pressure 760 mm hg
true (corrected) temperature of boiling water 100 °c
thermometer correction 1 °c
d. using the balance to calibrate your 10 ml pipet
temperature of water used in pipet 22.8 °c
corrected temperature.8 °c
trial 1 trial 2 trial 3
mass of erlenmeyer plus ~10 ml h₂o (gross mass) 47.55 g 39.05 g 40.55 g
mass of erlenmeyer (tare mass) 37.95 g 29.30 g 31.20 g
mass of ~10 ml of h₂o (net mass) 9.4 g 9.75 g 9.35 g

Explanation:

Step1: Calculate area of lab - book

We know length \(l = 10.8\) in and width \(w = 8.7\) in. First convert length and width to cm. Since \(1\) in \(= 2.54\) cm, length \(l=10.8\times2.54 = 27.432\) cm and width \(w = 8.7\times2.54=22.098\) cm. Area \(A=l\times w\), so \(A = 27.432\times22.098\approx606.1\) \(cm^{2}\). But if we use the values in cm from the report (\(l = 22.1\) cm and \(w = 7.5\) cm), then \(A=22.1\times7.5 = 165.75\approx166\) \(cm^{2}\).

Step2: Analyze thermometer correction

The observed temperature of boiling - water is \(99^{\circ}C\) and the true (corrected) temperature is \(100^{\circ}C\). The thermometer correction \(\Delta T=100 - 99=1^{\circ}C\).

Step3: Calculate mass of water in pipet trials

For trial 1: Mass of water \(m_1=\text{Mass of Erlenmeyer + water}-\text{Mass of Erlenmeyer}=47.55 - 37.95 = 9.6\) g. For trial 2: \(m_2=39.05 - 29.30 = 9.75\) g. For trial 3: \(m_3=40.55 - 31.20 = 9.35\) g.

Answer:

Area of lab - book: \(166\) \(cm^{2}\), Thermometer correction: \(1^{\circ}C\), Mass of water in trial 1: \(9.6\) g, Mass of water in trial 2: \(9.75\) g, Mass of water in trial 3: \(9.35\) g