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Afina-wow [57]
3 years ago
6

Which of the following is an accurate example of how the water cycle can influence local weather? A. An increase in evaporation

rates along a coastal area leads to afternoon thunderstorms. B. Water in a lake soaks into the local groundwater system. C. An increase in an area's precipitation causes the area to experience a drought. D. all of these
Chemistry
2 answers:
artcher [175]3 years ago
6 0

Answer:  A. An increase in evaporation rates along a coastal area leads to afternoon thunderstorms.

Explanation:

irina1246 [14]3 years ago
3 0

The answer is; A

During a hot day, the land heats up faster than the waters. The air on land becomes warm and less dense fast and begin to rise in the atmosphere. The air on the ocean with is still cooler and denser moves in to replace the rising on land air. This causes a sea breeze. The sea breeze carries with it, moisture. The hotter the day the higher the humidity. When the air goes inland, it causes precipitation when it rises,  cool, and condenses.


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jeyben [28]

Answer: A. is prey B. is predator C. is habitat D. is comunity E. is plants are producers and plant eating animals  consumors  F. is food chain G. is food web H. parasite I decomposers and scavengers J. is compete K an adsption

Explanation:

7 0
3 years ago
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sergeinik [125]

Answer:

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Explanation:

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6 0
3 years ago
Part A
Elden [556K]

Answer:

ΔG° = -5.4 kJ/mol

ΔG = 873.2 J/mol = 0.873 kJ /mol

Explanation:

Step 1: Data given

ΔG (NO2) = 51.84 kJ/mol

ΔG (N2O4)  = 98.28 kJ/mol

Step 2:

ΔG = ΔG° + RT ln Q

⇒with Q = the reaction quatient

⇒with T = the temperature = 298 K

⇒with R = 8.314 J / mol*K

⇒with ΔG° = ΔG° (N2O4) - 2*ΔG°(NO2 )

⇒ ΔG° = 98.28 kJ/mol - 2* 51.84  kJ/mol

⇒ ΔG° = -5.4 kJ/mol

Part B

ΔG =  ΔG° =RT ln Q

⇒with G° = -5.4 kj/mol = -5400 j/mol

⇒ with R = 8.314 J/K*mol

⇒with T = 298 K

⇒with Q = p(N2O4)/ [ p(NO2) ]² = 1.63/0.36² = 12.577

ΔG = -5400 + 8.314 * 298 * ln(12.577)

ΔG = -5400 + 8.314 * 298 * 2.532

ΔG = 873.2 J/mol = 0.873 kJ/mol

3 0
3 years ago
use the heisenberg uncertainty principle to calculate the uncertainity in meters in the position of 0.68g and traveling at a vel
My name is Ann [436]

Answer:

7.7439×10⁻³¹ m

Explanation:

The expression for Heisenberg uncertainty principle is:

\Delta x\times \Delta v=\frac {h}{4\times \pi\times m}

Where m is the mass of the microscopic particle

h is the Planks constant

Δx is the uncertainty in the position

Δv is the uncertainty in the velocity

Given:

mass = 0.68 g = 0.68×10⁻³ kg

Δv = 0.1 m/s

Δx= ?

Applying the above formula as:

\Delta x\times 0.1=\frac {6.62\times 10^{-34}}{4\times \frac {22}{7}\times 0.68\times 10^{-3}}

<u>Δx = 7.7439×10⁻³¹ m</u>

8 0
3 years ago
Which is a physical change?
ira [324]

Making toast is a physical change.



*** Please mark this answer as the Brainliest and leave a Thanks if I helped you! :) ***

4 0
3 years ago
Read 2 more answers
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