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2 years ago

When logging is carried out in a watershed, a likely effect on the local streams is.

Chemistry

1 answer:

guapka [62]2 years ago

4 0

Answer: A likely effect on the watershed resulting from logging would be increased water temperature.

Explanation: Not only will the removal of trees from the watershed would result in the decrease of Dissolved Oxygen(DO) in the water, but also increase temperature because the leaves of the trees absorb heat, which indirectly decreases the temperature of the river.

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Basile [38]

Answer:allow for greater chances for a sentinel event

Explanation:

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2 years ago

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[03.04]Based on the electronegativity values shown in the table below, which element would you expect to form a nonpolar covalen

vfiekz [6]

The answer is nitrogen c.

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3 years ago

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A weather balloon is inflated to a volume of 27.9L at a pressure of 732mmHg and a temperature of 30.1?C. The balloon rises in th

masya89 [10]

Answer:

45.4 L

Explanation:

Using Ideal gas equation for same mole of gas as

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

Given ,

V₁ = 27.9 L

V₂ = ?

P₁ = 732 mmHg

P₂ = 385 mmHg

T₁ = 30.1 ºC

T₂ = -13.6 ºC

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (30.1 + 273.15) K = 303.25 K

T₂ = (-13.6 + 273.15) K = 259.55 K

Using above equation as:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

$\frac{{732}\times {27.9}}{303.25}=\frac{{385}\times {V_2}}{259.55}$

$\frac{385V_2}{259.55}=\frac{20422.8}{303.25}$

$148225V_2=6729708.26018$

Solving for V₂ , we get:

V₂ = 45.4 L

7 0

3 years ago

The molecular structure of water contains two atoms of hydrogen and one atom of oxygen. When water reaches its boiling point and

nikklg [1K]

The water reaches its boiling point and turns into water vapor. This process is a physical change. The change is substance status. And the physical change will not destroy the molecular structure. So the molecular structure will not change.

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3 years ago

A 0.944 M solution of glucose, C6H12O6, in water has a density of 1.0624 g/mL at 20∘C. What is the concentration of this solutio

____ [38]

Answer:

Mole fraction: 0,0157.

Molality: 0,889m

Mass%: 16%

Explanation:

The units are mole fraction, molality and mass%

Units of mole fraction are moles of glucose per total moles.

Moles of glucose assuming 1L are:

0,944 moles.

Moles of water in 1L are:

1L × (1,0624kg/L) × (1000g / 1kg) × (1mol / 18,02g) = 59,0 moles of water

Mole fraction is: 0,944 moles / (59,0 mol + 0,944mol) = 0,0157

Molality is mole of solute (0,944) per kg of solution (1,0624kg):

0,944mol / 1,0624kg = 0,889m

In mass percent total mass is 1062,4g and mass of 0,944 moles of glucose is:

0,944mol×(180,156g/1mol) = 170g of glucose. Mass%:

170g / 1062,4g ×100 = 16%

I hope it helps!

6 0

3 years ago