We can observe physical properties of elements and compounds without changing the substance.
Examples of physical properties: Density, color, boiling point, state of matter, appearance: dull or shiny, etc.
But we can also observe and measure chemical properties by reacting a substance with something else. For example, like mixing baking soda and vinegar together. The vinegar reacts with the baking soda and produces carbon dioxide: a new substance.
Some examples of chemical properties: Flammability, amount of heat that is released during combustion, toxicity (how much damage it causes to other organisms), radioactivity, and ability to oxidize (when you have metal that becomes rusty looking).
The Importance of Monitoring<span> the </span>Tides<span> and Their Currents.Commercial and recreational fishermen use their knowledge of the </span>tides<span> and </span>tidal<span> currents to help them improve their catches. Depending on the species and water depth in a particular area, fish may concentrate during ebb or flood </span>tidal<span> currents.
Hope This Helps! :3</span>
Answer: Strong (b)
Explanation:
INFRARED ACTIVE BONDS
Not all covalent bonds display bands in the IR spectrum. Only polar bonds do so. These are referred to as IR active.
The intensity of the bands depends on the magnitude of the dipole moment associated with the bond in question:
Strongly polar bonds such as carbonyl groups (C=0) produce strong bands.
2Al(s) + 3Cl₂ → 2AlCl₃
Forming the equation:
2(28) + 3(223) = ΔH + 2(111)
ΔH = 503 kJ
Writing this outside of the reaction equation:
ΔH = -503 kJ
This uses the concept of freezing point depression. When faced with this issue, we use the following equation:
ΔT = i·Kf·m
which translates in english to:
Change in freezing point = vant hoff factor * molal freezing point depression constant * molality of solution
Because the freezing point depression is a colligative property, it does not depend on the identity of the molecules, just the number of them.
Now, we know that molality will be constant, and Kf will be constant, so our only unknown is "i", or the van't hoff factor.
The van't hoff factor is the number of atoms that dissociate from each individual molecule. The higher the van't hoff factor, the more depressed the freezing point will be.
NaCl will dissociate into Na+ and Cl-, so it has i = 2
CaCl2 will dissociate into Ca2+ and 2 Cl-, so it has i = 3
AlBr3 will dissociate into Al3+ and 3 Br-, so it has i = 4
Therefore, AlBr3 will lower the freezing point of water the most.
