Question

1. the table below shows some of the properties of two metals. a scientist has a sample of metal that could be either cobalt or nickel. which of the following properties could be tested to classify the metal sample? a. mass b. density c. electrical conductivity d. reactivity with oxygen 5. water is sometimes called the universal solvent because it is a polar molecule that dissolves many substances. which of the following is soluble in water? a. ice b. metals c. oil d. salts 6. gabriel is testing the properties of four - brands of golf balls. he places them in graduated cylinders containing 100 milliliters (ml) of water and determines that each ball has a volume of 20 cubic centimeters (cm³). he places the balls on a balance and finds that each has a different mass measured in grams (g). the properties of the golf balls are shown in the chart below. golf ball properties brand mass (g) volume (cm³) a 12 20 b 18 20 c 30 20 d 35 20 gabriel uses these numbers to find the density of the golf balls using this formula: density = mass/volume. water has a density of 1.0 g/cm³. which ball will float? a. ball a b. ball b c. ball c d. ball d

Explanation:

Step1: Recall solubility rules

Water is a polar solvent. Ionic compounds (salts) are generally soluble in water due to ion - dipole interactions. Ice is water in solid form, metals are insoluble as they have metallic bonding and oil is non - polar and insoluble in polar water.

Step2: Analyze each option

• Ice is water itself, not a solute being dissolved in water in the traditional sense of solubility.
• Metals are insoluble in water as they do not form favorable interactions with water molecules.
• Oil is non - polar and water is polar, so oil is insoluble in water due to the "like dissolves like" principle.
• Salts are ionic compounds and are soluble in water.

Step3: Recall density formula and floating condition

The density formula is $
ho=\frac{m}{V}$, where $
ho$ is density, $m$ is mass and $V$ is volume. A ball will float in water if its density is less than the density of water ($1.0\ g/cm^{3}$).

Step4: Calculate density of each ball

• For ball A: $

ho_A=\frac{m_A}{V_A}=\frac{22}{20}=1.1\ g/cm^{3}$

• For ball B: $

ho_B=\frac{m_B}{V_B}=\frac{18}{20}=0.9\ g/cm^{3}$

• For ball C: $

ho_C=\frac{m_C}{V_C}=\frac{30}{20}=1.5\ g/cm^{3}$

• For ball D: $

ho_D=\frac{m_D}{V_D}=\frac{33}{20}=1.65\ g/cm^{3}$

Step5: Determine floating ball

Ball B has a density of $0.9\ g/cm^{3}$ which is less than the density of water ($1.0\ g/cm^{3}$), so it will float.

Answer:

1. d
2. b