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babunello [35]
2 years ago
14

Explanation I need to show work pls help me

Chemistry
1 answer:
pishuonlain [190]2 years ago
5 0

Answer

D. Acid rain reacting with limestone bedrock.

Explanation:

- Brief Description:

  • A chemical reaction occurs when acidic rainfall falls on limestone or chalk. During the process, new, soluble compounds are produced. These disintegrate in the sea and are washed away, weathering the rock. Some forms of rock are resistant to chemical weathering. When rain falls from the sky onto a limestone (CaCO3) statue, a reaction between sulphuric acid and calcium carbonate happens. Calcium sulfate is formed as a result of this process (CaSO4). Because calcium sulfate is soluble in water, the statue will ultimately disintegrate. Limestone is chemically worn through the carbonation process. Rainwater collects carbon dioxide as it travels through the atmosphere, forming a weak carbonic acid. Water and carbon dioxide react to generate a mild carbonic acid. The fractures in the limestone are acted upon by this mild carbonic acid. Many weak acids, such as carbonic acid, are found in water. When carbon dioxide gas from the atmosphere combines with rainfall, a weak but copious acid is created. Other forms of acid rain are produced by sulfur dioxide and nitrogen gases, which function as chemical weathering agents.
<h3>- Chemical Weathering</h3>
  • Chemical weathering is generally the most active and effective weathering process. Water within soil or stone dissolves minerals of soil, softens minerals that absorpb the water, and dissolves carbondioxide.
  • Chemical weathering is the breakdown of rock by chemical processes. Water, air, and chemicals released by organisms cause chemical weathering of rocks when they dissolve the minerals in a rock.

- Carbon Dioxide-Bicarbonate-Carbonate Equilibrium

  • The carbon dioxide/bicarbonate/carbonate buffer is an essential buffer in surface waters. When water is in equilibrium with both CO2 from the atmosphere and carbonate-containing rock, its pH is buffered to 8.3, which is close to the pKa of the weak acid bicarbonate HCO3- (pKa = 8.4).
<h3>- Acid Rain </h3>
  • The production of nitric and sulfuric acids in our atmosphere causes acid rain. These chemicals are strong acids that are very soluble in water and dissolve in cloud water droplets.
  • The majority of nitrogen and sulfur oxides are caused by human activity. Electric utilities (60 percent), industrial combustion (17 percent), and industrial processes are the principal sources of sulfur dioxide emissions (8 percent ). Transportation, with internal combustion engines, accounted for more than half of all NOx emissions, with additional emissions from electric utilities (26 percent) and industrial combustion accounting for the remainder (14 percent ). Agricultural operations, particularly manure management, are the biggest source of ammonia emissions, but industry and transportation also emit some ammonia. Acid rain leads to the acidity of lakes and streams, as well as the degradation of trees at high elevations and vulnerable forest soils.
<h3>- Effect of Limestone</h3>
  • Calcium carbonate, often known as [Ca][CO3], is a common mineral. One well-known form of calcium carbonate is limestone. Acids in acid rain increase calcium carbonate breakdown by interacting with the carbonate anion.
  • This results in a bicarbonate solution. Because surface waters are in balance with atmospheric carbon dioxide, the concentration of carbonic acid, H2CO3, in the water remains constant.Because the minerals react with the excess acid, the presence of limestone and other calcium carbonate rock in lakes and streams helps to maintain a steady pH. However, acid rain can finally overwhelm the surface water's buffering ability.

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1. 0.240 liters of water would be needed to dissolve 21.6 g of lithium nitrate to make a 1.3 M (molar) solution.

2. 2.9 M is the molarity of a solution made of 215.1 g of HCl is dissolved to make 2.0 L of solution.

3.83.3 ml of concentrated 18 M H2SO4 is needed to prepare 250.0 mL of a 6.0 M solution.

4. 135 ml of stock HBr will be required to dilute the solution.

5. 150 ml of water should be added to 50.0 mL of 12 M hydrochloric acid to make a 4.0 M solution

6. The pH of the resulting solution is 13.89

Explanation:

The formula used in solving the problems is

number of moles= \frac{mass}{atomic mass of one mole}      1st equation

molarity = \frac{number of moles}{volume}            2nd equation

Dilution formula

M1V1 = M2V2          3rd equation

1. Data given

mass of Lithium nitrate = 21.6 grams

atomic mass of on emole lithium nitrate = 68.946 gram/mole

Molarity is given as 1.3 M

VOLUME=?

Calculate the number of moles using equation 1

n = \frac{21.6}{68.946}

  = 0.313 moles of lithium nitrate.

volume is calculated by applying equation 2.

volume = \frac{0.313}{1.3}

            = 0.240 litres of water will be used.

2. Data given:

mass of HCl = 215.1 gram

atomic mass of HCl = 36.46 gram/mole

volume = 2 litres

molarity = ?

using equation 1 number of moles calculated

number of moles = \frac{215.1}{36.46}

number of moles of HCl = 5.899 moles

molarity is calculated by using equation 2

M = \frac{5.899}{2}

   = 2.9 M is the molarity of the solution of 2 litre HCl.

3. data given:

molarity of H2SO4 = 18 M

Solution to be made 250 ml of 6 M

USING EQUATION 3

18 x V1= 250 x 6

V1 = 83.3 ml of concentrated 18 M H2SO4 will be required.

4. data given:

M1= 10M, V1 =?, M2= 3 ,V2= 450 ml

applying the equation 3

10 x VI = 3x 450

V1 = 135 ml of stock HBr will be required.

5. Data given:

V1 = 50 ml

  M1= 12 M

  V2=?

  M2= 4

applying the equation 3

50 x 12 = 4 x v2

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6. data given:

HCl + NaOH ⇒ NaCl + H20

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volume of NaOH = 25 ml

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volume of HCl = 0.335 ml

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Number of moles of NaOH and HCl is calculated by using equation 1 and converting volume in litres

moles of NaOH = 0.0131

moles of HCl= 0.025 moles

The ratio of moles is 1:1 . To find the unreacted moles of acid and base which does not participated in neutralization so the difference of number of moles of acid minus number of moles of base is taken.

difference of moles = 0.0119  moles ( NaOH moles is more)

Molarity can be calculated by using equation 1 in (25 +75 ml) litre of solution

molarity = \frac{0.0119}{0.1}

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14 = pH + pOH  

  pH  = 14 - 0.11

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