All categories

3 years ago

What geologic features allowed the Stovepipe Dunes to form?

2 answers:

4 0

Answer: Dunes are formed through the interaction of mountains and wind. The wind picks up quartz grains from the mountains in the area to form the dunes and these are reshaped by the same process over time.

damaskus [11]3 years ago

4 0

Answer:

<u>Sand dunes are geological features that are commonly formed in the areas where there is abundant concentration of sands</u>. This includes the bars in the rivers, deserts, and coastal beaches. The agent that is responsible for dune formation is wind. The sand size particles due to high wind get eroded from different places and transported to a place where the accumulation of sand is more.

It is a pile of sand and it has two sides, namely the wind-ward side, the direction from where the wind blows and the flip side, where there are no wind and sand grains deposited.

The stovepipe dunes in California have formed in this way.

You might be interested in

What is the area on the earth's surface directly above where an earthquake originates?

Natalija [7]

Answer:

The epicenter

5 0

3 years ago

Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1

Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

Explanation:

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of $SbCl_5$ is increasing .So, the equilibrium will shift in the right direction.

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

Concentration of $SbCl_5$ = 0.195 M

Concentration of $SbCl_3$ = $6.98\times 10^{-2} M$

Concentration of $Cl_2$ = $6.98\times 10^{-2} M$

On adding more $[SbCl_5$ to 0.370 M at equilibrium :

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

Initially

0.370 M $6.98\times 10^{-2}M$

At equilibrium:

(0.370-x)M $(6.98\times 10^{-2}+x)M$

The equilibrium constant of the reaction = $K_c$

$K_c=2.50\times 10^{-2}$

The equilibrium expression is given as:

$K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}$

$2.50\times 10^{-2}=\frac{(6.98\times 10^{-2}+x)M\times (6.98\times 10^{-2}+x)M}{(0.370-x) M}$

On solving for x:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

3 0

3 years ago

Mendeleev placed thallium (Tl) in the same group as lithium (Li), sodium (Na), potassium (K), rubidium (Rb), and cesium (Cs). Ho

EastWind [94]

Answer:

When observing how thallium reacts with the air of the earth's atmosphere, its hardness or resistance resembled sodium, it was not investigated further to classify it correctly

Explanation:

Now it is known that they contain different numbers of valence electrons and that thallium is a heavy metal like lead and that they have similar characteristics except for their melting point where thallium is higher.

4 0

2 years ago

Here is my question..

OverLord2011 [107]

I would say 3.0 cause yeah yeah yeah yeah I’m iann Dior

7 0

3 years ago

Read 2 more answers

What is the temperature of CO2 gas if the average speed (actually the root-mean-square speed) of the molecules is 750 m/s?

erastova [34]

Answer:

992.302 K

Explanation:

V(rms) = 750 m/s

V(rms) = √(3RT / M)

V = velocity of the gas

R = ideal gas constant = 8.314 J/mol.K

T = temperature of the gas

M = molar mass of the gas

Molar mass of CO₂ = [12 + (16*2)] = 12+32 = 44g/mol

Molar mass = 0.044kg/mol

From

½ M*V² = 3 / 2 RT

MV² = 3RT

K = constant

V² = 3RT / M

V = √(3RT / M)

So, from V = √(3RT / M)

V² = 3RT / M

V² * M = 3RT

T = (V² * M) / 3R

T = (750² * 0.044) / 3 * 8.314

T = 24750000 / 24.942

T = 992.302K

The temperature of the gas is 992.302K

Note : molar mass of the gas was converted from g/mol to kg/mol so the value can change depending on whichever one you use.

3 0

3 years ago