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3 years ago

At the end of seafloor spreading, where does the molten rock solidify?

Chemistry

1 answer:

olganol [36]3 years ago

8 0

Answer:

As the two plates move apart from each other, which often occurs at a rate of several centimetres per year, molten rock wells up from the underlying mantle into the gap between the diverging plates and solidifies into new oceanic crust. Spreading centres are found at the crests of oceanic ridges.

Explanation:

I found this answer by taking notes in class

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Benzene exists as a resonance hybrid and its bonds exhibit characteristics that are halfway between single and double bonds. Res

gregori [183]

Answer:

Explanation:

We can only talk about resonance hybrid for a compound in which more than one structure is possible based on its observed chemical properties.

There are compounds whose chemical properties can not be satisfactorily explained on the basis of a single chemical structure. In the case of such compounds, we invoke the idea of resonance.

A resonance hybrid is a single structure drawn to represent a given chemical specie which exhibits resonance behaviour and can otherwise be represented on paper in the form of an average of two or more chemical structures separated each from the next by a double-headed arrow.

5 0

3 years ago

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

3 years ago

I’ll give 5 stars and Brainliest plus this is 20 points lol

Daniel [21]

Answer:

I think is b

Explanation:

4 0

3 years ago

How many milliliters of sodium metal, with a density of 0.97 g/mL, would be needed to produce 53.2 grams of hydrogen gas in the

olchik [2.2K]

The balanced chemical reaction is:

<span>2Na + 2H2O → 2NaOH + H2
</span><span>
We first use the amount of hydrogen gas to be produced and the molar mass of the hydrogen gas to determine the amount in moles to be produced. Then, we use the relation from the reaction to relate H2 to Na.

53.2 g H2 ( 1 mol / 2.02 g ) ( 2 mol Na / 1 mol H2 ) ( 22.99 g / 1 mol ) = 1210.96 g Na

1210.96 g Na ( 1 mL / 0.97 g ) = 1248.41 mL Na needed</span>

7 0

3 years ago

What were Lamarck's ideas about evolution and why were those ideas incorrect

kow [346]

What were Lamarck's ideas about evolution and why were those ideas incorrect

4 0

3 years ago

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