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3 years ago

Which of the following factors does not affect soil formation? a. precipitation b. time c. erosion d. none of the above

Chemistry

2 answers:

Keith_Richards [23]3 years ago

7 0

Answer: Option (a) is the correct answer.

Explanation:

Precipitation is defined as the formation of a solid in a liquid which is not soluble.

Therefore, presence of an insoluble solid substance does not affect soil formation.

Whereas when natural forces like wind, water etc which erode away rocks and soil are known as erosion. Also, time is a factor that also affects the soil formation.

For example, in rainy season due to heavy rain soil erosion occurs which affects the soil formation.

laiz [17]3 years ago

3 0

the answer is a,) because it particpates a substance from a solution. hope this helps

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A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

Romashka [77]

Answer: The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

Explanation:

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

Calculation for freezing point of solution:

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

Calculation for boiling point of solution:

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

8 0

3 years ago

PLEASE HELP!!! PLEASE.

adell [148]

Answer:

Q₁: [HCl] = 0.075 N = 0.075 M.

Q₂: [KOH] = 7.675 mN = 7.675 mM.

Q₃: [H₂SO₄] = 0.2115 N = 0.105 M.

Q₄: The equivalence point is the point at which the added titrant is chemically equivalent completely to the analyte in the sample whereas the endpoint is the point where the indicator changes its color.

Explanation:

Q₁: If it takes 67 mL of 0.15 M NaOH to neutralize 134 mL of an HCl solution, what is the concentration of the HCl?

As acid neutralizes the base, the no. of gram equivalent of the acid is equal to that of the base.

The normality of the NaOH and HCl = Their molarity.

∵ (NV)NaOH = (NV)HCl

∴ N of HCl = (NV)NaOH / (V)HCl = (0.15 N)(67 mL) / (134 mL) = 0.075 N.

∴ The concentration of HCl = 0.075 N = 0.075 M.

Q₂: If it takes 27.4 mL of 0.050 M H₂SO₄ to neutralize 357 mL of KOH solution, what is the concentration of the KOH solution?

As mentioned in Q1, the no. of gram equivalent of the acid is equal to that of the base at neutralization.

The normality of H₂SO₄ = Molarity of H₂SO₄ x 2 = 0.050 M x 2 = 0.1 N.

∵ (NV)H₂SO₄ = (NV)KOH

∴ N of KOH = (NV)H₂SO₄ / (V)KOH = (0.1 N)(27.4 mL) / (357 mL) = 7.675 x 10⁻³ N = 7.675 mN.

∴ The concentration of KOH = 7.675 mN = 7.675 mM.

Q₃:If it takes 55 mL of 0.5 M NaOH solution to completely neutralize 130 mL of sulfuric acid solution (H₂SO₄), what is the concentration of the H₂SO₄ solution?

As mentioned in Q1 and 2, the no. of gram equivalent of the acid is equal to that of the base at neutralization.

The normality of NaOH = Molarity of NaOH = 0.5 N.

∵ (NV)H₂SO₄ = (NV)NaOH

∴ N of H₂SO₄ = (NV)NaOH / (V)H₂SO₄ = (0.5 N)(55 mL) / (130 mL) = 0.2115 N.

∴ The concentration of H₂SO₄ = 0.2115 N = 0.105 M.

Q₄: Explain the difference between an endpoint and equivalence point in a titration.

The equivalence point is the point at which the added titrant is chemically equivalent completely to the analyte in the sample whereas the endpoint is the point where the indicator changes its color.

The equivalence point in a titration is the point at which the added titrant is chemically equivalent completely to the analyte in the sample. It comes before the end point. At the equivalence point, the millimoles of acid are chemically equivalent to the millimoles of base.

End point is the point where the indicator changes its color. It is the point of completion of the reaction between two solutions.

The effectiveness of the titration is measure by the close matching between equivalent point and the end point. pH of the indicator should match the pH at the equivalence to get the same equivalent point as the end point.

6 0

3 years ago

In the scientific method, analysis should follow right after

alina1380 [7]

Observation/ question
research
hypothesis
experiment
analysis
conclusion

7 0

3 years ago

In the pie chart below, which of the following represents the amount of Earth's water located in the oceans?

Andrew [12]

Answer:

D. 97%

Explanation:

♪＼(*＾▽＾*)／＼(*＾▽＾*)／

4 0

3 years ago

Read 2 more answers

What molarity is prepared when a solution contains 0.277 moles of calcium bromide in 500 mL of solution (must change mL to Liter

Vladimir79 [104]

Answer:

0.554M of Calcium Bromide

Explanation:

Molarity by defintion is #of moles of something/litres of solution.

Therefore, here, we have 0.277 moles of calcium bromide and 500mL (divide 500ml by 1000 to go from mL to L because for every 1L there's 1000mL) or 0.5L.

Molarity= 0.277/0.5 = 0.554M of Calcium Bromide

5 0

3 years ago