Answer:Artificial light from cities has created a permanent "skyglow" at night, obscuring our view of the stars. Here's their map of artificial sky brightness in North America, represented as a ratio of "natural" nighttime sky brightness. In the black areas, the natural night sky is still (mostly) visible.
Explanation:
Answer:
See explanation
Explanation:
The reactivity of metals has a lot to do with their position in the electrochemical series. However, it is also known that metallic character decreases across the period. This implies that as we move from left to right along the periodic table. Sodium, magnesium, aluminum and silicon continues to decrease in metallic character. As a matter of fact, silicon is a metalloid and not a pure metal.
Sodium reacts with cold water to give a vigorous reaction,magnesium and aluminium reacts with steam at red heat.
Silicon does not react with water, even as steam, under normal conditions.
Answer:
positive charge
Explanation:
Protons are positively charged
The correct answer is option b, that is, triggered by changes in the weather.
Hibernation refers to a state of lower metabolism and inactivity usually witnessed in endotherms. This condition is featured by a slow heart rate, slow breathing, low metabolic rate, and low body temperature. It assists the majority of the endotherms to thrive in the cold seasons. For example, hibernation of a bear in winters. Therefore, weather change can stimulate hibernation.
Answer:
81°C.
Explanation:
To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat released from water (Q = - 1200 J).
m is the mass of the water (m = 20.0 g).
c is the specific heat capacity of water (c of water = 4.186 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = final T - 95.0°C).
∵ Q = m.c.ΔT
∴ (- 1200 J) = (20.0 g)(4.186 J/g.°C)(final T - 95.0°C ).
(- 1200 J) = 83.72 final T - 7953.
∴ final T = (- 1200 J + 7953)/83.72 = 80.67°C ≅ 81.0°C.
<em>So, the right choice is: 81°C.</em>
