Uncertainty of the data hope its help you
Answer:
Option E. Zirconium
Explanation:
From the question given above, the following data were obtained:
Length of side (L) of cube = 0.2 cm
Mass (m) of cube = 52 mg
Name of the unknown metal =?
Next, we shall determine the volume of the cube. This can be obtained as follow:
Length of side (L) of cube = 0.2 cm
Volume (V) of the cube =?
V = L³
V = 0.2³
V = 0.008 cm³
Next, we shall convert 52 mg to g. This can be obtained as follow:
1000 mg = 1 g
Therefore,
52 mg = 52 mg × 1 g / 1000 mg
52 mg = 0.052 g
Thus, 52 mg is equivalent to 0.052 g.
Next, we shall determine the density of the unknown metal. This can be obtained as follow:
Mass = 0.052 g.
Volume = 0.008 cm³
Density =?
Density = mass / volume
Density = 0.052 / 0.008
Density of the unknown metal = 6.5 g/cm³
Comparing the density of the unknown metal i.e 6.5 g/cm³ with those given in table in the above, we can conclude that the unknown metal is zirconium
Explanation:
The mass of a pot is 300g and contains 90% aluminum. Find the number of moles of aluminum in the pot. P.A. (Al = 27)
The mass of aluminum present in the pot is:
Hence, in the given pot 270g Al is present.
The gram atomic mass of Al -27 g/mol
Given the mass of Al is 270 g
Substitute these values in the above formula:
Answer is 10.0 mol of Al is present.
Answer:
Here are a few more examples:
Smoke and fog (Smog)
Dirt and water (Mud)
Sand, water and gravel (Cement)
Water and salt (Sea water)
Potassium nitrate, sulfur, and carbon (Gunpowder)
Oxygen and water (Sea foam)
Petroleum, hydrocarbons, and fuel additives (Gasoline)
Heterogeneous mixtures possess different properties and compositions in various parts i.e. the properties are not uniform throughout the mixture.
Examples of Heterogeneous mixtures – air, oil, and water, etc.
Examples of Homogeneous mixtures – alloys, salt, and water, alcohol in water, etc.
Explanation:
