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Juli2301 [7.4K]
3 years ago
15

When scientists looked at the polarity of bands of rock on either side of the mid-ocean ridges, what did they find? A. that ther

e was no pattern B. that the patterns were different C. that the polarities were at right angles D. that the patterns of polarity matched up on both sides
Chemistry
2 answers:
storchak [24]3 years ago
8 0

Answer:

B. that the patterns were different

Explanation:

  • When the scientist looked at the oceanic ridges at the surface they found on either side were the polarity of bands of rocks as on either side of the oceanic ridges where the positive and negative anomalies.  
  • Being composed of hard and soft rocks, these tend to have been in strips of black and white bands or rocks aligned due to the upwelling of new lava on the sides of the ridges throughout the evolutionary history of the planet.
  • Polarities have also been documented to have been aligned in coordination with earth magnetic poles, Being basaltic these rocks have varying ages as older rocks are found away from the mid-oceanic ridges due to the impact of spreading out of lava or magma.
alisha [4.7K]3 years ago
3 0
I am pretty sure the answer is D. That the patterns of a polarity matched up on both sides.
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Explanation:

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vampirchik [111]

Answer: The percent yield for the NaBr is, 86.7 %

Explanation : Given,

Moles of FeBr_3 = 2.36 mol

Moles of NaBr = 6.14 mol

First we have to calculate the moles of NaBr

The balanced chemical equation is:

2FeBr_3+3Na_2S\rightarrow Fe_2S_3+6NaBr

From the reaction, we conclude that

As, 2 moles of FeBr_3 react to give 6 moles of NaBr

So, 2.36 moles of FeBr_3 react to give \frac{6}{2}\times 2.36=7.08 mole of NaBr

Now we have to calculate the percent yield for the NaBr.

\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield = 6.14 moles

Theoretical yield = 7.08 moles

Now put all the given values in this formula, we get:

\text{Percent yield}=\frac{6.14mol}{7.08mol}\times 100=86.7\%

Therefore, the percent yield for the NaBr is, 86.7 %

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3 years ago
How much heat (in kJ) would need to be removed to cool 150.3 g of water from 25.60°C to -10.70°C?
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Q = -22.9 kJ

Explanation:

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