Question

1. write the following numbers in standard notation: (4 pts)

a. 5.01 x 10^-2
b. 3.102 x 10^6

1. convert the following measurements. (8 pts)

a. 292 k to °c
b. 36°f to °c
c. 0.06 tons to grams
d. 9,511 ml to quarts
concept 3:

1. the marble in the figure to the right has a mass of 3.75 g. when placed in a graduated cylinder the water level changes, as pictured. assume the graduated cylinder is measuring in ml. if water has a density of 0.997 g/ml, will the marble float or not in the graduated cylinder? justify your answer with evidence and reasoning. (2 pts)
2. write a sentence using the terms atom, element, and compound to describe propane, which has the following chemical formula: c₃h₈. (3 pts)
3. differentiate between a compound and a mixture. (2 pts)

Explanation:

28.

Step1: For \(a = 5.01\times10^{- 2}\), move decimal 2 places left

\(5.01\times10^{-2}=0.0501\)

Step2: For \(b = 3.102\times10^{6}\), move decimal 6 places right

\(3.102\times10^{6}=3102000\)

Step1: Convert Kelvin to Celsius (\(T_{C}=T_{K}-273.15\)) for \(a\)

\(T_{C}=292 - 273.15=18.85^{\circ}C\)

Step2: Convert Fahrenheit to Celsius (\(T_{C}=\frac{5}{9}(T_{F}-32)\)) for \(b\)

\(T_{C}=\frac{5}{9}(36 - 32)=\frac{5}{9}\times4=\frac{20}{9}\approx2.22^{\circ}C\)

Step3: Convert tons to grams (\(1\) ton \( = 907185\) g) for \(c\)

\(0.06\times907185 = 54431.1\) g

Step4: Convert milliliters to quarts (\(1\) quart \(=946.353\) mL) for \(d\)

\(9511\div946.353\approx10.05\) quarts

Step1: Find density of marble (\(

ho=\frac{m}{V}\)). First find \(V\) by water - level change. Assume initial water level \(V_1\) and final \(V_2\), \(V = V_2 - V_1\). Let's assume \(V_2 - V_1=4\) mL (from the graduated - cylinder reading). Given \(m = 3.75\) g.
\(
ho_{marble}=\frac{3.75}{4}=0.9375\) g/mL

Step2: Compare with water density

Since \(
ho_{marble}=0.9375\) g/mL and \(
ho_{water}=0.997\) g/mL, and \(
ho_{marble}<
ho_{water}\)

Answer:

a. \(0.0501\)
b. \(3102000\)

29.