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Butoxors [25]
3 years ago
9

Which sequence of events is required to form a limestone cave where you can walk around and observe cave formations, such as sta

lactites?
Chemistry
1 answer:
N76 [4]3 years ago
5 0

Answer:

Which sequence of events is required to form a limestone cave where you can walk around and observe cave formations, such as stalactites? (Note: stalactites hang from the ceiling - they have to hold on tight to the roof.)

A geological sequence of events as involving the lowering of the water table to expose cave structures where stalactites and stalagmites form which is described as follows,

Explanation:

1. Acidic percolated water formed cavities of solution beneath the natural water table known as phreatic zone

2. After the passage of time there is a drop in the water table dropped forming caves from cavities

3. These caves, which are air filled voids that contains adequate environment for forming stalactites and stalagmites and where they are found

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A sample of nitrogen gas is produced in a reaction and collected under water in a graduated cyliner. The temperature is 26.3 oC
Rudiy27

The mass of nitrogen collected is mathematically given as

M-N2=0.025gram

<h3>What is the mass of nitrogen collected?</h3>

Question Parameters:

A sample weighing 2.000g

the liberated NH3 is caught in  50ml pipeful  of H2SO4 (1.000ml   =  0.01860g Na2O).

T=26.3c=299.3K

Pressure=745mmHg=745torr

Pressure of N2=745-25.2=719.8torr

Generally, the equation for the ideal gas   is mathematically given as

PV=nRT

Therefore

719.8/760=45.6/1000=n*0.0821*299.3

n=0.00176*14

In conclusion, the Mass of N2

M-N2=0.00176*14

M-N2=0.025gram

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8 0
2 years ago
¿que se necesita para escribir un informe de experimento? Plis ​
Ludmilka [50]

Answer:

Explanation:

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4 0
2 years ago
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Salt is poured from a container at 10 cm³ s-¹ and it formed a conical pile whose height at any time is 1/5 the radius of the abo
Romashka-Z-Leto [24]

Answer:

\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}

Explanation:

Volume of a cone:

  • \displaystyle V=\frac{1}{3} \pi r^2 h

We have \displaystyle \frac{dV}{dt} = \frac{10 \ cm^3}{sec} and we want to find \displaystyle \frac{dh}{dt} \Biggr | _{h\ =\ 6}= \ ? when the height is 2 cm.

We can see in our equation for the volume of a cone that we have three variables: V, r, and h.

Since we only have dV/dt and dh/dt, we can rewrite the equation in terms of h only.

We are given that the height of the cone is 1/5 the radius at any given time, 1/5r, so we can write this as r = 5h.

Plug this value for r into the volume formula:

  • \displaystyle V =\frac{1}{3} \pi (5h)^2 h  
  • \displaystyle V =\frac{1}{3} \pi \ 25h^3

Differentiate this equation with respect to time t.

  • \displaystyle \frac{dV}{dt}  =\frac{25}{3} \pi \ 3h^2 \ \frac{dh}{dt}
  • \displaystyle \frac{dV}{dt}  =25 \pi h^2 \ \frac{dh}{dt}

Plug known values into the equation and solve for dh/dt.

  • \displaystyle 10 = 25 \pi (2)^2  \ \frac{dh}{dt}
  • \displaystyle 10 = 100 \pi  \ \frac{dh}{dt}  

Divide both sides by 100π to solve for dh/dt.

  • \displaystyle \frac{10}{100 \pi} = \frac{dh}{dt}
  • \displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}

The height of the cone is increasing at a rate of 1/10π cm per second.

7 0
3 years ago
A 33.6 mL solution of KCl has a mass of 27.29 g. After evaporating to dryness, the dry salt residues has a mass of 6.31 g. Calcu
bixtya [17]

For  the molarity and the  percentage of the mass of kcl are mathematically given as

%m/m = 23.1%, molarity  = 3.38M

<h3>What is the molarity?</h3>

Generally, the equation for the  percentage of the mass of kcl is mathematically given as

(mass of kcl/mass of solution)x 100

therefore

%m/m = (6.31/27.29)x100

%m/m = 23.1%

The molarity

(no of mol/ vol of sol)

molarity  = 0.0846mol/0.0250 L

molarity  = 3.38M

In conclusion, the molarity is given as

molarity  = 3.38M

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4 0
2 years ago
The solution in the two arms of the U-tube are separated by a membrane that is permeable to water and glucose but not to sucrose
Gnom [1K]

Answer:

The liquid level will rise in Side A and drop in Side B.

Step-by-step explanation:

The membrane is impermeable to sucrose, but permeable to sucrose and water.

<u>      Side A   </u>    <u>    Side B    </u>  

2 M sucrose   1 M sucrose

1 M glucose   2 M glucose

(a) Ignoring osmotic effects

The glucose will diffuse spontaneously from the side with higher concentration to that of lower concentration until equilibrium is established. There is no change in volume on either side.

At this point, we have

<u>     Side A       </u>   <u>        Side B   </u>      

2    M sucrose    1 M sucrose

1.5 M glucose    1.5 M glucose

=====

(b) With osmotic effects

The solute concentration on Side A is greater than on Side B.

Water will diffuse into Side A.

The liquid level will rise in Side A and drop in Side B.

8 0
3 years ago
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