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Andru [333]
3 years ago
8

the curved movement of air or water is the result of which of these. A).Winds speeding off equater B).the rotation of earth on i

ts axis C.)the different air movements north or south of the equater or d).the friction that slows down winds near earths surface.
Chemistry
2 answers:
Ivahew [28]3 years ago
8 0
The curved movement of air or water is the result of which of these:
The answer is B
Aleksandr-060686 [28]3 years ago
8 0

B. The rotation of earth on it's axis

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Active sites in enzymes are where substrates bind. Once they bind, a catalytic reaction occurs as a complex between substrate and enzyme is formed. Enzymes are termed as biocatalysts or simply catalysts since they speed up the catalytic reaction. After the reaction, they simply revert back to their original form, after having adjusted to fit with substrate. 
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3 years ago
PLEASE HELP WILL REWARD BRAINLIEST
riadik2000 [5.3K]
The answer is the principal quantum number

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7 0
3 years ago
Read 2 more answers
What’s IV DV and Controls
zavuch27 [327]

Answer:

Iv-Independent Variable the part that doesnt get affected by the Dependent. DV- Dependent Variable- the part that gets affected by the independent. Controls is what everything is gettong compared to

7 0
3 years ago
Calculate the freezing point of a solution that contains 8.0 g of sucrose (C12H22O11) in 100 g of H2O. Kf for H2O = 1.86C/m
Aleks04 [339]

The freezing point of the sucrose solution is -0.435°C.

<h3>What is the freezing point of the solution?</h3>

The freezing point of the solution is determined from the freezing point depression formula below:

  • ΔT = mKf(H₂O)

Kf(H₂O) = 1.86 Cm

m is molality of solution = moles of solute/mass of solvent

moles of sucrose = 8.0/342.3 = 0.0233 moles

m = 0.0233/0.1 = 0.233 molal

ΔT = 0.233 m * 1.86°C/m.

ΔT = 0.435 °C.

Freezing point of sucrose solution = 0°C - 0.435°C

Freezing point of sucrose solution  = -0.435°C.

In conclusion, the freezing point of sucrose solution is determined from the freezing point depression.

Learn more about freezing point depression at: brainly.com/question/19340523

#SPJ1

8 0
2 years ago
If the pressure on a gas at -73°C is doubled but its volume is held constant, what will its final temperature be in degrees Cels
Gre4nikov [31]

Answer:

127°C

Explanation:

This excersise can be solved, with the Charles Gay Lussac law, where the pressure of the gas is modified according to absolute T°.

We convert our value to K → -73°C + 273 = 200 K

The moles are the same, and the volume is also the same:

P₁ / T₁ = P₂ / T₂

But the pressure is doubled so: P₁ / T₁ = 2P₁ / T₂

P₁ / 200K = 2P₁ / T₂

1 /2OOK = (2P₁ / T₂) / P₁

See how's P₁ term is cancelled.

200K⁻¹ = 2/ T₂

T₂ = 2 / 200K⁻¹  → 400K

We convert the T° to C → 400 K - 273 = 127°C

4 0
3 years ago
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