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3 years ago

Which currents are deflected by the eastern coastline of the continents?

Chemistry

1 answer:

kotegsom [21]3 years ago

6 0

Answer:

The correct answer is option B, that is, western boundary currents.

Explanation:

The Western boundary currents get deflected by the eastern coastline of the continents. The western boundary currents refer to the deep, warm, narrow, and fast flowing currents, which produce on the west end of the ocean basins because of the western intensification. They conduct warm water from the tropics poleward. Some of the examples of western boundary currents are the Agulhas current, Gulf Stream, and the Kuroshio.

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How much does the earth weigh?

Lady_Fox [76]

Answer:

5.972 × 10^24 kg

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3 years ago

Examples of the skills used in science include _____ and _____.

ollegr [7]

In science you want to use observations and measurements.

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3 years ago

A mixture of gases contains 4.46 moles of neon (Ne), 0.74 mole of argon (Ar), and 2.15 moles of xenon (Xe). Calculate the partia

viva [34]

Explanation:

The partial pressure of an individual gas is equal to the total pressure of the mixture multiplied by the mole fraction of the gas.

Total pressure = 2atm

Mole Fraction = number of moles / total number of moles

Neon

Mole Fraction = 4.46 / 7.35 = 0.607

Partial Pressure = 0.607 * 2 = 1.214 atm

Argon

Mole Fraction = 0.74 / 7.35 = 0.101

Partial Pressure = 0.101 * 2 = 0.202 atm

Xenon

Mole Fraction = 2.15 / 7.35 = 0.293

Partial Pressure = 0.293 * 2 = 0.586 atm

5 0

3 years ago

Solid Al(NO3)3 is added to distilled water to produce a solution in which the concentration on nitrate, [NO3^-], is 0.10 M. What

VikaD [51]

Answer:

The concentration of nitrate ion $NO_{3}^{-}$ is 0.033 M.

Explanation:

$Al(NO_{3})_{3} \rightarrow Al^{3+} + NO_{3}^{-}$

1 mole of $[Al^{3+}]$ = 3 mole $NO_{3}^{-}$

Let $[NO_{3}^{-}]$ = 3y

then according to above reaction,

$[Al^{3+}]$ = y

$[NO_{3}^{-}]$ = 0.10 M (given)

$[Al^{3+}]$ = ?

3y = 0.10 M

$y = \frac{0.1}{3}$

y = 0.033 M

So ,concentration of $[Al^{3+}]$ = y = 0.033 M

5 0

3 years ago

Calculate the Kelvin temperature to which 10.0 L of a gas at 27 °C would have to be heated to change the volume to 12.0 L. Units

KatRina [158]

Hello!

For this problem, we will be applying Charles' Law:

V1/T1 = V2/T2

Now that we have the formula, let's convert the temperature to Kelvin.

27 + 273 = 300K

Let's plug everything in now!

10/300 = 12.0/x

Simplified:

1/30 = 12.0/x

Cross-multiply:

1x = 30*12.0

x = 360

Check!

10/300 = 12/360

300*12 = 360*10

3600 = 3600

Therefore, you would have to heat the gas at a temperature of 360K in order to raise the volume to 12.0L.

4 0

3 years ago