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igor_vitrenko [27]
3 years ago
15

What role does heat energy from the sun play influencing ocean currents

Chemistry
1 answer:
alexandr1967 [171]3 years ago
5 0

The sun heats up the atmosphere at the equator. The dense water sinks under and forms deep water masses under the ocean current. Then it gets wind due to the air current and this brings currents at the surface if the ocean.

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How many moles are in 7.1x10^21 atoms of iron?
Nitella [24]

Answer:

0.011 moles

Explanation:

There are about 6.02*10^23 atoms in a mole, so in the given sample, there are

\frac{7.01 \times  {10}^{21} }{6.02 \times  {10}^{23} }

which is about 0.011 moles.

8 0
2 years ago
What is isostasy?
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7 0
2 years ago
1
Molodets [167]

Answer: The temperature does effect the growth of the plants. If you kept them all at different temperature and the growth was different then the temperature is definitly a factor in plant growth

Explanation:

7 0
3 years ago
Which sample of matter is classified as a substance?<br> (1) air (3) milk(2) ammonia (4) seawater
jeyben [28]
A chemical substance has the characteristics that it cannot be separated by physical  methods. Seawater and milk can be separated by sedimentation, and air has different components depending on other aspects (such as elevation). Only ammonia is a substance. (thus it can have a formula: NH<span>3)</span>
3 0
2 years ago
Calculate the mass (grams) of 0.750 moles of Al2(Cr2O7)3
garik1379 [7]

Answer:

526g is the mass of this sample

Explanation:

To solve this question we must, as first, find the <em>molar mass </em>of Al₂(Cr₂O₇)₃ using the periodic table. The molar mass is defined as the mass of this compound per mole. With this value we can find the mass in 0.750 moles as follows:

<em>Molar mass Al₂(Cr₂O₇)₃</em>

2Al = 2*26.98g/mol = 53.96g/mol

6 Cr = 6*51.9961g/mol = 311.9766g/mol

21 O = 21*15.999g/mol = 335.979g/mol

53.96g/mol + 311.9766g/mol + 335.979g/mol

= 701.9156g/mol

The mass of 0.750 moles is:

0.750 moles * (701.9156g / mol) =

<h3>526g is the mass of this sample</h3>
8 0
2 years ago
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