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2 years ago

Stalactites and stalagmites are examples of _____.

Chemistry

2 answers:

aleksandr82 [10.1K]2 years ago

5 0

Answer:

The correct answer is option D, chemical sedimentary rock

Explanation:

Both stalactites and stalagmites are called drip stone as they are formed of the dripping water consisting of minerals. Both stalactites and stalagmites hang up from the ceiling of caves. These are also refereed to as Chemical sedimentary rocks formed by precipitation. The three major processes involved in the formation of these two rocks are –

a) Physical weathering – which disintegrates the minerals contained in the igneous, sedimentary, or metamorphic rock

b) Chemical weathering – which involves dissolution of weathered rock minerals

c) Precipitation of the dissolved minerals.

professor190 [17]2 years ago

3 0

The right option is; chemical sedimentary rock

Chemical sedimentary rocks are rocks that are formed by precipitation of minerals from water. Chemical sedimentary rocks are also called evaporates and they are usually composed of the minerals halite (calcium chloride) and gypsum (calcium sulfate). Examples of chemical sedimentary rocks include stalactites, stalagmites, rock salt, flint, and some dolomites.

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3 Ni2+(aq) + 2 Cr(OH)3(s) + 10 OH− (aq) → 3 Ni(s) + 2 CrO42−(aq) + 8 H2O(l) ΔG∘ = +87 kJ/molGiven the standard reduction potenti

Mrrafil [7]

Answer:

The standard reduction potential E°cell (Cr6+/Cr3+) is -0.13V

Explanation:

Step 1: Data given

3 Ni^2+(aq) + 2 Cr(OH)3(s) + 10 OH− (aq) → 3 Ni(s) + 2 CrO4^2−(aq) + 8 H2O(l) ΔG∘ = +87000 J/mol

Ni2+(aq) + 2 e− → Ni(s) E∘red = -0.28 V

Step 2: The half reactions:

Cathode: Ni2+(aq) + 2 e− → Ni(s) E° = -0.28 V

Anode: CrO4^2-(aq) + 4H2O(l) +3e- → Cr(OH)3(s) + 5OH- (aq) E°= unknown

Step 3: Calculate E°cell

ΔG° = -n*F*E°cell

⇒ with ΔG° = the gibbs free energy

⇒ n = the number of electrons in the net reaction = 6

⇒ F = the Faraday constant = 96485 C

⇒ E°cell= the standard cell potential

Step 4: Calculate E°(Cr6+/Cr3+

E°cell= ΔG°/(-n*F)

E°cell = 87000 /(-6*96485)

E°cell = -0.150 V

E°cell = E°(Ni2+/Ni) - E°(Cr6+/Cr3+)

E°(Cr6+/Cr3+) = -0.13V

The standard reduction potential E°cell (Cr6+/Cr3+) is -0.13V

7 0

2 years ago

Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic

Kisachek [45]

Answer:

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C) It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

$\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9$

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = $164.5\times 10^{-9} g$

$1 ng=10^{-9} g$

Volume of the sample = V = $15.3 cm^3$

Density of the lake water sample ,d= $1.00 g/cm^3$

Mass of sample = M = $d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g$

$ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75$

10.75 is the concentration of arsenic in the sample in parts per billion.

Mass of arsenic in $1 cm^3$ of lake water = $\frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g$

Mass of arsenic in $0.710 km^3$ lake water be m.

$1 km^3=10^{15} cm^3$

Mass of arsenic in $0.710\times 10^{15} cm^3$ lake water :

$m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g$

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

$\frac{2.74}{41.90} days$

Then days required to remove 7,633.66 kg of arsenic from the lake water :

$=7,633.66\times \frac{2.74}{41.90} days=499.19 days$

1 year = 365 days

499.19 days = $\frac{499.19}{365} years = 1.367 years\approx 1.37 years$

It will take 1.37 years to remove all of the arsenic from the lake.

3 0

2 years ago

If you perform the experiment described in investigation #15 by mixing 10 g of glue with 13 g of water and 8 g of sodium borate

Phantasy [73]

10 g of glue with 13 g of water ,

Mass ratio of the material can be calculated as:

$\frac{Mass of glue}{Mass of glue + Mass of water}$

$\frac{10 g}{10 g + 13 g }$

$\frac{10 g}{23 g }$

8 g of sodium borate suspended in 11 g of water, mass ratio can be calculated as:

$\frac{Mass of sodium borate}{Mass of sodium borate + Mass of water}$

$\frac{ 8 g}{ 8 g + 11 g }$

$\frac{8 g}{19 g }$

3 0

2 years ago

The gas in a sealed container has an absolute pressure of 125.4 kilopascals. If the air around the container is at a pressure of

AlexFokin [52]

Answer: C. 25.6 kPa

Explanation:

The Gauge pressure is defined as the amount of pressure in a fluid that exceeds the amount of pressure in the atmosphere.

As such, the formula will be,

PG = PT – PA

Where,

PG is Gauge Pressure

PT is Absolute Pressure

PA is Atmospheric Pressure

Inputted in the formula,

PG = 125.4 - 99.8

PG = 25.6 kPa

The gauge pressure inside the container is 25.6kPa which is option C.

4 0

2 years ago

How many grams o an ointment base must be added to 45 g o clobetasol (T EMOVAT E) ointment, 0.05% w/w, to change its strength to

saw5 [17]

Answer:

Drug calculation

If we have 45g of clobetasol = 0.05%w/w

Then what mass in g of clobetasol is in 0.03%w/w = 45 x 0.03/0.05 =27g

It means that 27g of clobetasol must be added to change the drug strength to 0.03% w/w

3 0

2 years ago