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ehidna [41]
3 years ago
9

Which of the following provides evidence for past tectonic plate motions?

Chemistry
2 answers:
irinina [24]3 years ago
4 0

Answer:

so i am doing this and i need help what did u get for number 1

Explanation:

its the first question

Ronch [10]3 years ago
3 0

Answer:

D

Explanation:

If there are 2 of the same rocks on two continents, that shows that they were once merged. If there is glacial existence in places where glaciers would’ve melted (Like Africa has glacial existence from Antarctica ) this shows that continents were once merged. If life forms preserved as fossils were matched on 2 different continents this shows that all the continents were once merged.

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Please help... I am begging...
iVinArrow [24]

Answer:

The number, such as 98.7 FM, of a radio station represents:

  • <u>the frequency in which is transmitted the radio signal</u>.

Explanation:

<em>The radio FM is the modulated frequency, which means that all the information is sent by just a signal, with different frequencies which difference them</em>, <u>the radio FM use the frequencies from 88 MHz until 108 Mhz (MHz is a measuring unit for the frequency), with a minimal space among them of 0.2 MHz</u>, this last means that you could find a signal in 88.0 MHz, and the next should be 88.2 MHz, next 88.4 MHz and so (at least, regularly the space between two frequencies is more than 0.2 MHz).

5 0
4 years ago
Technetium-104 has a half life of 18.0 minutes. how much of a 165.0 g sample remains after 90.0 minutes?
11Alexandr11 [23.1K]

The mass of 165.0 g sample that remains after 90.0 minutes is 5.16 grams

calculation

lambda㏑2/18= 0.0385

m(t)= 165 x e( 0.0385 x90)  =5.16g

8 0
3 years ago
Read 2 more answers
1 mol super cooled liquid water transformed to solid ice at -10 oC under 1 atm pressure.
Arada [10]

Answer:

Explanation:

Given that:

number of moles of super cooled liquid water = 1

Melting enthalpy of ice = 6020 J/mol

Freezing point =0 °C = (0 + 273 K)= 273 K

The decrease in entropy of the system during freezing for 1 mol (i.e during transformation from liquid water to solid ice )  = - 6020 J/mol × 1 mol /273 K = -22.051 J/K

Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)

\Delta \ S = \int\limits^{T_2}_{T_1}\dfrac{nC_p(s)dT}{T}

\Delta \ S = {nC_p(s)In \dfrac{T_2}{T_1}

\Delta \ S = {(1*37.7)In \dfrac{263}{273}

Δ S = -1.4 J/K

Total entropy change of the system = - 22.05 J/K - 1.4 J/K = - 23.45 J/K

Entropy change of universe = entropy change of the system+ entropy change of the surrounding

According to the second law of thermodynamics

Entropy change of universe  >0

SO,

Entropy change of the system + entropy change in the surrounding > 0

Entropy change in the surrounding > - entropy change of the system

Entropy change in the surrounding > - (- 23.53 J/K)

Entropy change in the surrounding > 23.53 J/K

b) Make some comments on entropy changes from the obtained data.

From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.

4 0
3 years ago
The air pressure inside a submarine is 1.24 atm. What would be the height of a column of mercury (mm Hg) balanced by this pressu
Ipatiy [6.2K]
Since one atmosphere is 29.9213 inches of mercury, we can simply find this by taking 29.9213 and multiplying that by 1.24. You end up with 37.102412 inches of mercury. I hope this helps you out, and good luck on the rest of your homework.
6 0
4 years ago
Maple syrup, which comes from the sap of maple trees, is a mixture of water and natural sugars. It's a clear, brown liquid. It's
Margaret [11]
Correct Answer: Option C i.e <span>Solution

Reason: 
Solutions are characterized by particles of size less than 1nm. Since the particle size in solutions are very small, they cannot be separated by centrifugation. On other hand, colloids have particle size ranging from 1nm to 100 nm, while suspensions have particle size > 100 nm. Hence, they can be separated by centrifugation. </span>
5 0
4 years ago
Read 2 more answers
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