The two substances that are mostly likely examples of covalent bonding are Sucrose and Ethanol.
<h3 /><h3 /><h3>What is a covalent Bond?</h3>
- A covalent bond is a type of chemical bond that involves the sharing of pairs of electron between atoms.
Examples of compounds with covalent bond include the following;
- Distilled water
- Sucrose
- Ethanol
Olive oil is a mixture not a compound
Sodium Chloride & Potassium lodide are examples of ionic bond.
Thus, the two substances that are mostly likely examples of covalent bonding are Sucrose and Ethanol.
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Answer:
Cheetah cubs are in danger from predators like lions and hyenas which can track their prey by scent and so the mother and her cubs leave an area when their scent is too strong so that they are not hunted and the cubs survive.
Mother Cheetahs also train their cubs to hunt so that they may get food for themselves which will ensure their survival as well thus showing that both of these practices can impact on reproductive success.
The efficiency of a light source is the percentage of its energy input that gets radiated as visible light if some of the blue light in an led is used to cause a fluorescent material to glow the overall efficiency of the LED decreases.
How efficient is LED?
Different wavelengths that correlate to different visible colours are used in LED light therapy. Various shades pierce the skin at different rates.
- Your skin's outermost layer is impacted by blue light.
- Yellow light is more enveloping.
- Red light penetrates your skin more deeply.
- The deepest penetrating light is near-infrared.
Different LED hues have various effects. For instance, according to experts red LED light therapy has the potential to reduce inflammation and boost collagen formation, which declines with age and is crucial for maintaining youthful-looking skin.
Acne-causing bacteria may be destroyed by a blue LED light therapy (P. acnes).
Hence the answer is the overall efficiency of the LED decreases.
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D. 51 N. The minimum applied force that will cause the television slide is 51 N.
In order to solve this problem we have to use the force of static friction equation Fs = μs*n, where μs is the coefficient of static friction, and n is the normal force m*g.
With μs = 0.35, and n = 15kg*9.8m/s² = 147 N
Fs = (0.35)(147 N)
Fs = 51.45 N
Fs ≅ 51 N