The number of years required for 1/4 cobalt-60 to remain after decay is calculated as follows
after one half life 1/2 of the original mass isotope remains
after another half life 1/4 mass of original mass remains
therefore if one half life is 5.3 years then the years required
= 2 x 5.3years = 10.6 years
The answer is gas only.
hope this helps!
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:
Answer:
B. Solvent
Explanation:
In osmosis, water always moves from low solute concentration to high solute concentration. SOLUTE NEVER MOVES AS IT CANNOT PASS THE SELECTIVELY PERMEABLE MEMBRANE. alot of caps but need to stress this concept cuz otherwise this concept gets very confusing
