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2 years ago

What makes a cyclone rotate counter-clockwise north of the Equator and clockwise south of the Equator?

Chemistry

2 answers:

julia-pushkina [17]2 years ago

8 0

The Coriolis Effect (Core-nol-e-us)

saw5 [17]2 years ago

5 0

The ocean currents move differently on both equator‘s so it makes the cyclones rotate different

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You just got home from a run on a hot Atlanta afternoon. You grab a 1.00 liter bottle of water and drink three quarters of it in

-Dominant- [34]

Answer:

¾ litters of water was consumed

2.24 litters = 1 mole of water

Therefore ¾÷2.24 moles was consumed

0.33 moles

8 0

3 years ago

A buffer solution contains 0.345 M acetic acid and 0.377 M sodium acetate . If 0.0613 moles of potassium hydroxide are added to

melamori03 [73]

Answer:

pH = 5.54

Explanation:

The pH of a buffer solution is given by the Henderson-Hasselbach (H-H) equation:

pH = pKa + log $\frac{[CH_3COO^-]}{[CH_3COOH]}$

For acetic acid, pKa = 4.75.

We calculate the original number of moles for acetic acid and acetate, using the given concentrations and volume:

CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻

CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH

The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.

Now we use the H-H equation to calculate the new pH, by using the new concentrations:

pH = 4.75 + log $\frac{(0.0942+0.0613)mol/0.250L}{(0.0862-0.0613)mol/0.250L}$ = 5.54

6 0

3 years ago

The average weather conditions of a region

Verdich [7]

Answer:

A) Climate

Hope this helps !

6 0

3 years ago

Look at sample problem 19.10 in the 8th ed Silberberg book. Write the Ksp expression. Find the concentrations of the ions you ne

777dan777 [17]

Answer:

Explanation:

From the information given:

$CaF_2 \to Ca^{2+} + 2F^-$

$Ksp = 3.2 \times 10^{-11}$

no of moles of $Ca^{2+}$ = 0.01 L × 0.0010 mol/L

no of moles of $Ca^{2+}$ = $1 \times 10^{-5} \ mol$

no of moles of $F^-$ = 0.01 L × 0.00010 mol/L

no of moles of $F^-$ = $1 \times 10^{-6}\ mol$

Total volume = 0.02 L

$[Ca^{2+}}] = \dfrac{1\times10^{-5} \ mol}{0.02 \ L} \\ \\ \\ \[[Ca^{2+}}] = 0.0005 \ mol/L$

$[F^{-}] = \dfrac{(1\times 10^{-6} \ mol)}{0.02 \ L}$

$[F^{-}] = 5 \times 10^{-5} \ mol/L$

$Q = [Ca^{2+}][F^-]^2 \\ \\ Q = 0.0005 \times (5\times 10^{-5})^2 \\ \\ Q = 1.25 \times 10^{-12}$

Since Q<ksp, then there will no be any precipitation of CaF2

3 0

3 years ago

A sample of methane gas occupies 4.50 gallons at a temperature of 80.0 F. If the pressure is held constant, what will be the vol

Marat540 [252]

Answer:

21.19 L or 5.6 gallons

Explanation:

Data Given

initial volume V1 = 4.50 gallons

convert volume from gallons to L

1 gallon = 3.78541 L

4. 50 gallon = 3.785 L x 4.50

So,

initial volume V1 = 17.03 L

final Volume V2 = ?

initial Temperature T1 = 80° F

convert Temperature to Kelvin

T1 = ( 80° F -32) / 1.8 +273

T1 = 26.7° C + 273 = 299.7 K

final Temperature T2 = 100° C

convert Temperature to Kelvin

T2 = 100° C + 273 = 373 K

Solution:

This problem will be solved by using charles' law equation at constant pressure.

The formula used

V1 / T1 = V2 / T2

As we have to find out volume, so rearrange the above equation

V2 = V1 x T2 / T1

Put value from the data given and conversion information in above equation

V2 = 17.03 L x 373 K / 299.7 K

V2 = 21. 19 L

So the final volume of Methane gas = 21.19 L

Or for Volume in gallons

1 L = 0.26

21.19 x 0.264 = 5.6 gallons

7 0

3 years ago