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4 years ago

________ are formed when groundwater dissolves cavities into limestone.

Chemistry

2 answers:

beks73 [17]4 years ago

8 0

Answer:

Caves.

Explanation:

Hello,

In this case, caves are formed by the dissolution of limestone. It turns out that rainwater picks up carbon dioxide from the air and as it percolates through the soil it turns into a weak acid (carbonic acid). This causes a slowly dissolution of the limestone along the joints, bedding planes and fractures, some of which become enlarged enough to form caves.

Best regards.

SpyIntel [72]4 years ago

6 0

The answer is Caverns

as Caverns of limestone is a natural cavity, which forms in the subsurface of the earth by the effect of acidic groundwater and underground water (like rivers) making the limestone dissolve by time and leaves this cavity. which increase and grow up with time.
and there are many structures which form in this caverns ex :
(Speleothem, Stalactite, Flowstone, Stalagmite, Helictite)

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3 years ago

What is the mass, in grams, of a liquid having a density of 1.50g/mL and a volume of 3500mL

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Answer:

5,250g

Explanation:

Density = Mass / Volume

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3 years ago

Hello, everyone!

marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the law of ideal gases,

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x $\frac{1 g}{1000 mg}$ x $\frac{1 mol}{28.01 g}$

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V = $\frac{0.00125 mol x 0.082057 \frac{atm L}{mol K} x 243 K}{0.92 atm}$

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x $\frac{1000 cm^{3} }{1 L}$ = 27.09 cm³

Then the acceptable concentration c of CO in ppm is,

c = 27 cm³ / m³ = 27 ppm

To express this concentration in percent by volume we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 $\frac{cm^{3} }{m^{3} }$ x $\frac{1 m^{3} }{1 000 000 cm^{3} }$ x 100%

c = 0.003 %

So, the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is 27.09 ppm and 0.003 %.

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