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Svetradugi [14.3K]
2 years ago
13

I give brainliest

Chemistry
2 answers:
Tatiana [17]2 years ago
8 0

Answer: During winter the Northern Hemisphere leans away from the sun, there are fewer daylight hours and the sun hits us at an angle. This makes it appear lower in the sky. In equatorial regions the length of days and the directness of sunlight don't change as much.

Hope this helps :)

Citrus2011 [14]2 years ago
3 0

Answer:

What he said... more light gives more energy to plants, making them grow more.

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2 years ago
Click on the question mark. Choose the MOST reasonable inference based on the text evidence and what you already know. "in a anc
katrin2010 [14]

Answer:

in an ancient

Explanation:

when the word begins with a vowel you say an instead of a

8 0
3 years ago
Which group on the periodic table would have zero electronegativity because they have a full octet?
bija089 [108]
The group on the periodic table that would have 0 electronegativity due to the fact that their valence shell is full, i.e, have a full octet would be the inert or noble gases. They have a total of 8 electrons in their valence shell and are thus inert and cannot strongly attract electrons toward itself, from neighbouring atom electrons as it does not need to.
3 0
3 years ago
It took batman 5 minutes to catch robin in the bat-mobile. If he traveled 45 km before catching him, how fast was he going?
Nataly_w [17]

Answer:

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7 0
2 years ago
Consider the reaction: CaCO3(s) à CaO(s) + CO2(g)
Vsevolod [243]

Answer:

131.5 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

First, we will calculate the standard enthalpy of the reaction (ΔH°).

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g) ) - 1 mol × ΔH°f(CaCO₃(s) )

ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)

ΔH° = 179.2 kJ

Then, we calculate the standard entropy of the reaction (ΔS°).

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g) ) - 1 mol × S°(CaCO₃(s) )

ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)

ΔS° = 160.2 J/K = 0.1602 kJ/K

Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.

ΔG° = ΔH° - T × ΔS°

ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K

ΔG° = 131.5 kJ

6 0
3 years ago
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