Chemistry, 27.01.2021 19:50 jphan19

A landscape in Antarctica has an area of 500,000 square miles. It is covered with ice, giving the landscape an average elevation of 7520 feet. Without the ice, it is estimated the landscape would have an average elevation of 1500 ft. The average density of the ice is 0.917 g/mL. How many tonnes of ice cover that landscape?

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Answer from: Brayner14

m = 2.18x10¹⁵ tons of ice

Explanation:

To solve this exercise, we need to work with the same units of conversion. In this case let's see the conversion of every unit here:

1 mi = 5280 ft

1 ton = 1000 kg

1 kg = 1000 g

1 mL = 1 cm³

1 mi = 1609.34 m or 160,934 cm

Now that we know the units, let's analyze the data.

The landscape has a 500,000 mi² of area, this is however without including the ice. The ice is apart, and has a 7520 ft of height. And if the landscape wouldn't have ice, then the height will be 1500 ft only.

With that data, we can know the actual height of the ice by a simple difference:

Hi = Hli - Hl

Where:

Hi: Height of ice

Hli: Height of ice + landscape

Hl: Height of landscape

Applying this expression, the height of ice is:

Hi = 7520 - 1500 = 6020 ft of ice

Now that we know the height of the ice, we can calculate the volume of whole landscape. But first, let's do the conversion of foot to mile:

Hi = 6020 ft * (1 mi/5280 ft) = 1.14 mi

And now, let's calculate the volume of the landscape in mile, and then, to centimeter:

V = 1.14 mi * 500,000 mi² = 570,000 mi³

To convert the cubic mile to cubic centimeter we just do the following:

V = 570,000 mi³ * (160934 cm / 1 mi)³ = 2.38x10²¹ cm³

Now, we can get the mass of the ice using the given density of ice (d = 0.917 g/cm³)

If d = m/V

then m = d * V

Replacing the data:

m = 0.917 g/cm³ * 2.38x10²¹ cm³

m = 2.18x10²¹ g

Finally, the tons of ice will be:

m = 2.18x10²¹ g * (1 kg / 1000 g) * (1 ton / 1000 kg)

m = 2.18x10¹⁵ tons of ice

Hope this helps

Answer from: Quest

For cyclohexane compounds drawn in a chair conformation: -axial bonds are vertical -equatorial bonds are parallel to two bonds within the ring. thus, as i assume the chair conformation is drawn: -the x group is in an axial position, as are the five other vertical bonds. -three axial bonds are on the top face of the ring -the other three axial bonds are on the bottom face the position of the second substituent on each ring carbon is equatorial. gauche groups must be on an adjacent carbon such as carbon 6 and carbon 2, and also gauche groups are rotated by 60 degrees relative to each other. drawing the newman projections for c1–c2 and c1–c6 aids in visualizing the groups that are gauche to the y group on c1. click this link and watch some academic videos that may you more understand the topic

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