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3 years ago

Which seismic wave refracts and cannot penetrate the core

2 answers:

8 0

Recordings of seismic waves from earthquakes led to the discovery of the earth's core and eventual maps of the layers of the Earth's inside. Just as the prism below refracts light at its faces, seismic waves bend, reflect and change speed at the boundaries between different materials below the Earth's surface.

goldfiish [28.3K]3 years ago

8 0

Answer: P waves and S waves travel through the interior of the earth. P waves are the Primary waves and S waves are the Secondary waves.

P waves can travel in both solid as well as in liquid (outer core) whereas S wave can travel only through solid. When P waves enters the outer core, it gets deviated and again enters the outer core with further deviation.

In case of S waves, once it reaches the core-mantle boundary, it reflects to the side of it, forming a shadow zone commonly known as the S wave shadow zone (103°-103°). Thus it cannot penetrate through the outer core.

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Appropriate units for the speed of a chemical reaction, the reaction rate, are _______.

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Appropriate units for the speed of a chemical reaction, the reaction rate, are M/s

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3 years ago

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choli [55]

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Explanation:

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vovikov84 [41]

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3 years ago

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of water is pro

sergeinik [125]

The question is incomplete, here is the complete question:

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water. If 12.5 g of water is produced from the reaction of 72.6 g of sulfuric acid and 77.0 g of sodium hydroxide, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it.

Answer: The percent yield of water in the reaction is 46.85 %.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For NaOH:

Given mass of NaOH = 77.0 g

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaOH}=\frac{77.0g}{40g/mol}=1.925mol$

For sulfuric acid:

Given mass of sulfuric acid = 72.6 g

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 1, we get:

$\text{Moles of sulfuric acid}=\frac{72.6g}{98g/mol}=0.741mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

1 mole of sulfuric acid reacts with 2 moles of NaOH

So, 0.741 moles of sulfuric acid will react with = $\frac{2}{1}\times 0.741=1.482mol$ of NaOH

As, given amount of NaOH is more than the required amount. So, it is considered as an excess reagent.

Thus, sulfuric acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of sulfuric acid produces 2 moles of water

So, 0.741 moles of sulfuric acid will produce = $\frac{2}{1}\times 0.741=1.482moles$ of water

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 1.482 moles

Putting values in equation 1, we get:

$1.482mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.482mol\times 18g/mol)=26.68g$

To calculate the percentage yield of water, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of water = 12.5 g

Theoretical yield of water = 26.68 g

Putting values in above equation, we get:

$\%\text{ yield of water}=\frac{12.5g}{26.68g}\times 100\\\\\% \text{yield of water}=46.85\%$

Hence, the percent yield of water in the reaction is 46.85 %.

8 0

3 years ago

The following pairs of soluble solutions can be mixed. In some cases, this leads to the formation of an insoluble precipitate. D

Gekata [30.6K]

Answer:

See explanation

Explanation:

Let us see what happens when each solution is mixed;

a) AlCl3(aq) + K3PO4(aq) ------> 3KCl(aq) + AlPO4(s)

A precipitate is formed here

b) RbCO3(aq) + NaCl(aq) -------> This is an impossible reaction hence no solid precipitate is formed here

c) MnCl2(aq) + Na2CO3(aq) → 2NaCl(aq) + MnCO3(s)

A precipitate is formed.

d) K2S(aq) + 2NH4Cl(aq) ------> 2KCl(aq) + (NH4)2S(aq)

No solid precipitate is formed

e) CaCl2(aq) + (NH4)2CO3(aq) → CaCO3(s) + 2NH4Cl(aq)

A solid precipitate is formed

3 0

3 years ago