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DENIUS [597]
3 years ago
5

What is a cone of depression and how is it caused?

Chemistry
2 answers:
Nutka1998 [239]3 years ago
4 0
A cone of depression occurs in an aquifer when groundwater is pumped from a well. In an unconfined aquifer (water table), this is an actual depression of the water levels. In confined aquifers (artesian), the cone of depression is a reduction in the pressure head surrounding the pumped well.

When a well is pumped, the water level in the well is lowered. By lowering this water level, a gradient occurs between the water in the surrounding aquifer and the water in the well. Because water flows from high to low water levels or pressure, this gradient produces a flow from the surrounding aquifer into the well.

As the water flows into the well, the water levels or pressure in the aquifer around the well decrease. The amount of this decline becomes less with distance from the well, resulting in a cone-shaped depression radiating away from the well. This, in appearance, is similar to the effect one sees when the plug is pulled from a bathtub. This conical-shaped feature is the cone of depression.
Alja [10]3 years ago
3 0

Answer:

Depression can be caused by many things, like being neglected, abused, or even trauma. Even thoughts that might trigger sadness, or anxiety it's caused by alot of problems of your life or others around you

Explanation:

hope this helps a bit :)

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Please help!!! Is it balanced or unbalanced?
Black_prince [1.1K]

Hey there!

I believe the answer you are looking for is Unbalanced!

If I am incorrect I am very sorry and would love any feedback.

Have a nice day, and Happy Halloween!

6 0
3 years ago
A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb
andre [41]

Answer:

m_{PbI_2}=18.2gPbI_2

Explanation:

Hello,

In this case, we write the reaction again:

Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2}  *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI}  *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2

Best regards.

5 0
3 years ago
Read 2 more answers
What stress would shift the equilibrium position of the following system to the left?
Margarita [4]

Answer:

Heating the system

Explanation:

According to the principle of Le Chatelier, for a system at equilibrium, a specific disturbance would make the equilibrium shift toward the direction which minimizes such a disturbance.

Since we wish to shift the equilibrium to the left, this means we wish to increase the concentration of products, as an excess in their concentration would make the products react and produce more reactants in order to lower the excess concentration of products.

Since heat is also a product, an increase in heat would shift the equilibrium toward the left, as this would consume the excess of heat by producing the reactants.

4 0
3 years ago
if an object has a mass of 60 grams in a volume of 120 cm3 then calculate the density would this object sink or float
Ivanshal [37]

Answer: 0.5 g/cm^3

Density equals mass divided by volume so..

60/120 is 0.5 g/cm^3

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