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3 years ago

Logging is a cause of nonpoint source pollution how does the removal of trees affect our fresh water supply

2 answers:

8 0

Logging may cause the soil to be less compact and the loose dirt will enter and contaminate any water source.

Have a nice day!

kenny6666 [7]3 years ago

7 0

Answer Choices :

A: The removal of trees does not affect our fresh water supply.

B: Tree roots actually help absorb some chemicals, which prevents them from entering our fresh water supply.

C: Fewer trees increases absorption of water and other chemicals into the groundwater supply.

D: Fewer trees increases erosion, which adds more sediment to our fresh water supply.

Now Mr.Josh-Ray Says "Logging may cause the soil to be less compact and the loose dirt will enter and contaminate any water source."

Meaning - The Loose Dirt Contains Particles Are Contaminating Clean Water Sources.

D - Fewer trees increases erosion, which adds more sediment to our fresh water supply.

Therefore Mr.Josh Is Correct

"D" IS CORRECT ON PLATO.

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How many moles of nitrogen are contained in 18.65 L at STP?

stira [4]

1 mole of any gas under STP ----- 22.4 L

18.65 L*1 mol/22.4 L ≈ 0.8326 mol N2

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3 years ago

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Which best describes the current atomic theory?

Dennis_Churaev [7]

Atoms are composed of electrons in a cloud around a positive nucleus.

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The second-order rate constants for the reaction of oxygen atoms witharomatic hydrocarbons have been measured (R. Atkinson and J

Pepsi [2]

Answer:

A = 1,13x10¹⁰

Ea = 16,7 kJ/mol

Explanation:

Using Arrhenius law:

ln k = -Ea/R × 1/T + ln(A)

You can graph ln rate constant in x vs 1/T in y to obtain slope: -Ea/R and intercept is ln(A).

Using the values you will obtain:

y = -2006,9 x +23,147

As R = 8,314472x10⁻³ kJ/molK:

-Ea/8,314472x10⁻³ kJ/molK = -2006,9 K⁻¹

Ea = 16,7 kJ/mol

Pre-exponential factor is:

ln A = 23,147

A = e^23,147

A = 1,13x10¹⁰

I hope it helps!

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3 years ago

Mashutka [201]

sodium cloride is salt created from sodium Na and chlorine Ci

Na-sodium Ca- calcium

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H- hydrogen He-helium

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3 years ago

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A 2.350×10−2 M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving a sample of NaCl in water and then b

sveta [45]

Answer:

Part A: m = 0.02356 mol/kg = 0.02356 m

Part B: Xsolute = 4.243×10⁻⁴

Part C: % m/m = 0.1376%

Part D: ppm = 1,376 ppm

Explanation:

1. Data:

a) M = 2.350×10⁻² M

b) V sol = 1.000 L

c) V H₂O = 994.4 mL = 0.9944 L

d) d H₂O = 0.9982 g/mL

2. Formulae

M = n solute / V sol (L)
m = n solute / Kg solvent
X solute = n solute / N total
% m/m = (mass of solute / mass of solution) × 100
ppm = (mass of solute / mass of solution) × 1,000,000
density = mass in grams / volume in mL

3. Solution

Part A: Calculate the molality of the salt solution.

m = n solute / Kg solvent

i) M = n solute / V sol (L) ⇒ n solute = M × V sol (L)

⇒ n solute = M = 2.350×10⁻² M × 1.000 L = M = 2.350×10⁻² mol

ii) density H₂O = mass H₂O / volume H₂O

⇒ mass H₂O = density H₂O × volume H₂O

⇒ mass H₂O = 0.9982 g/mL × 999.4 mL = 997.6 g

iii) kg H₂O = 997.6 g / (1,000 g/Kg) = 0.9976 kg

iv) m = 2.350×10⁻² mol / 0.9976 kg = 0.02356 mol/kg = 0.02356 m

Part B: Calculate the mole fraction of salt in this solution.

X solute = n solute / N total

i) n solute = 2.350×10⁻² mol

ii) n solvent = n H₂O = mass H₂O in grams/ molar mass H₂O

⇒ 997.6 g / 18.015 g/mol = 55.38 mol

iii) X solute = 2.350×10⁻² mol / 55.38 mol = 4.243×10⁻⁴

Part C: Calculate the concentration of the salt solution in percent by mass.

% m/m = (mass of solute / mass of solution) × 100

i) molar mass = mass in grams / molar mass

⇒ mass of solute = mass of NaCl = n solute × molar mass NaCl

⇒ mass of solute = 2.350×10⁻² mol × 58.44 g/mol = 1.373 g

ii) % m/m = (1.373 g / 997.6 g) × 100 = 0.1376%

Part D: Calculate the concentration of the salt solution in parts per million.

ppm = (mass of solute / mass of solution) × 1,000,000

i) ppm = ( (1.373 g / 997.6 g) × 1,000,000 = 1,376 ppm

5 0

3 years ago