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3 years ago

What is the contour interval for this map? Scale : 1:24,000in/in PLEASE HELP!!!!

Chemistry

1 answer:

Nimfa-mama [501]3 years ago

4 0

Answer: A

Explanation:
A contour interval is a vertical distance or difference in elevation between contour lines. Index contours are bold or thicker lines that appear at every fifth contour line.

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________________ in the atmosphere and ocean creates wind and ocean currents.

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3 years ago

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How do molecular shape affect polarity

devlian [24]

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1 year ago

A solid yellow stripe is to be painted in the middle of a certain highway. If 1 gallon of paint covers an area of p square feet

Zolol [24]

Answer:

440mt/p gallons

Explanation:

We are told that one gallon of paint covers an area of p square feet. Knowing this then, the question becomes how much paint is needed to cover a stripe that is t inches wide and m miles long.

Let us first use the dimensions of the stripe to find how much area that is in square feet. A stripe is rectangular so we will calculate the area as follows:

Area of stripe = width x length

= t inches x m miles (we need to convert both to feet)

= (t inches x 1 ft/12 inches) x (m miles x 5280 ft/1 mi)

=440mt square feet

ow we can find out how much paint will be needed by using ratios

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6 0

3 years ago

A solution of 2-propanol and 1-octanol behaves ideally. Calculate the chemical potential of 2-propanol in solution relative to t

andrew-mc [135]

Answer:

The chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.

Explanation:

The chemical potential of 2-propanol in solution relative to that of pure 2-propanol can be calculated using the following equation:

$\mu (l) = \mu ^{\circ} (l) + R*T*ln(x)$

Where:

μ (l): is the chemical potential of 2-propanol in solution

μ° (l): is the chemical potential of pure 2-propanol

R: is the gas constant = 8.314 J K⁻¹ mol⁻¹

T: is the temperature = 82.3 °C = 355.3 K

x: is the mole fraction of 2-propanol = 0.41

$\mu (l) = \mu ^{\circ} (l) + 8.314 \frac{J}{K*mol}*355.3 K*ln(0.41)$

$\mu (l) = \mu ^{\circ} (l) - 2.63 \cdot 10^{3} J*mol^{-1}$

$\mu (l) - \mu ^{\circ} (l) = - 2.63 \cdot 10^{3} J*mol^{-1}$

Therefore, the chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.

I hope it helps you!

8 0

3 years ago