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lana [24]
3 years ago
10

Explain how the processes that occur at mid-ocean ridges and ocean trenches illustrate the cycling of matter.

Chemistry
1 answer:
Nimfa-mama [501]3 years ago
3 0

Explanation:

At one end, new crusts are being produced, at other end the crust is being destroyed and this strikes a unique balance.

At the mid-ocean ridges, the lithospheric plates are diverging. This is implies that the earth is pulling apart here. When the earth is pulling apart, new materials from the asthenosphere comes to the surface thereby creating new lithospheric plate.

As new plates are formed, they push back against the old one. New plates are found very close to the margin and it begins to age away from the margin.

On the other end, old plates are taken away from this center to ocean trenches. At oceanic trenches subduction is occurring.

In a subduction, the lithospheric plate plunges deep into the asthenosphere where they are being melted back.

This is a covergent margin.

This process continues in a dynamic manner to cycle matter on earth.

learn more:

Sea floor spreading brainly.com/question/9912731

#learnwithBrainly

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Zinc can be removed from bronze by placing bronze in hydrochloric acid. The zinc reacts with the hydrochloric acid producing zin
Ber [7]

Answer:The Zinc Reacts With The Hydrochloric Acid Producing Zinc Chloride And Hydrogen Gas, And Leaving The Copper Behind. A. If 25.0 G Of Zinc ... Zn+ 2 HCI --> ZnCl2 + H2 (answer .771 G H2) B. If The Reaction Yields . ... If 25.0 g of zinc are in a sample of bronze, determine the theoretical yield of hydrogen gas. Zn+ 2 HCI

8 0
3 years ago
Which element is less electronegative than silicon (Si)? A. Sulfur (S) B. Magnesium (Mg) C. Carbon (C) D. Oxygen (0)​
Luden [163]

Answer:

C. Carbon

Explanation:

Carbon has an electronegativity of 2.55, followed by Tin at 1.96, Silicon at 1.90 and the least electronegative would be Lead at 1.87.

7 0
2 years ago
Help. chemistry...........
nasty-shy [4]
<span>The </span>equilibrium<span> will </span>shift<span> to favor the side of the reaction that involves fewer moles of gas.
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8 0
3 years ago
Read 2 more answers
Calculate the standard enthalpy change for the reaction at 25 ∘C. Standard enthalpy of formation values can be found in this lis
weqwewe [10]

Answer:

\Delta H_{rxn}=-2043.999kJ

Explanation:

\Delta H_{rxn}^{0}=\sum [n_{i}\times \Delta H_{f}^{0}(product)_{i}]-\sum [n_{j}\times \Delta H_{f}^{0}(reactant_{j})]

Where n_{i} and n_{j} are number of moles of product and reactant respectively (equal to their stoichiometric coefficient).

\Delta H_{f}^{0} is standard heat of formation and \Delta H_{rxn}^{0} is standard enthalpy change for reaction at 25^{0}\textrm{C}

So, \Delta H_{rxn}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]

or, \Delta H_{rxn}=[3mol\times -393.509kJ/mol]+[4mol\times -241.818kJ/mol]-[1mol\times -103.8kJ/mol]-[5mol\times 0kJ/mol]

or, \Delta H_{rxn}=-2043.999kJ

4 0
3 years ago
Given the following reaction: 2K3PO4 + AL2(CO3)3 = 3K2CO3 + 2ALPO4 If I perform this reaction with 150 g of potassium phosphate
Novosadov [1.4K]

Answer : The theoretical yield of potassium carbonate is, 146.483 g

The percent yield of potassium carbonate is, 85.33 %

Solution : Given,

Mass of K_3PO_4 = 150 g

Mass of Al_2(CO_3)_3 = 90 g

Molar mass of K_3PO_4 = 212.27 g/mole

Molar mass of Al_2(CO_3)_3 = 233.99 g/mole

Molar mass of K_2CO_3 = 138.205 g/mole

First we have to calculate the moles of K_3PO_4 and Al_2(CO_3)_3

\text{ Moles of }K_3PO_4=\frac{\text{ Mass of }K_3PO_4}{\text{ Molar mass of }K_3PO_4}=\frac{150g}{212.27g/mole}=0.7066moles

\text{ Moles of }Al_2(CO_3)_3=\frac{\text{ Mass of }Al_2(CO_3)_3}{\text{ Molar mass of }Al_2(CO_3)_3}=\frac{90g}{233.99g/mole}=0.3846moles

The given balanced reaction is,

2K_3PO_4+Al_2(CO_3)_3\rightarrow 3K_2CO_3+2AlPO_4

From the given reaction, we conclude that

2 moles of K_3PO_4 react with 1 mole of Al_2(CO_3)_3

0.7066 moles of K_3PO_4 react with \frac{1}{2}\times 0.7066=0.3533 moles of Al_2(CO_3)_3

But the moles of Al_2(CO_3)_3 is, 0.3846 moles.

So, Al_2(CO_3)_3 is an excess reagent and K_3PO_4 is a limiting reagent.

Now we have to calculate the moles of K_2CO_3.

As, 2 moles of K_3PO_4 react to give 3 moles of K_2CO_3

So, 0.7066 moles of K_3PO_4 react to give \frac{3}{2}\times 0.7066=1.0599 moles of K_2CO_3

Now we have to calculate the mass of K_2CO_3.

\text{ Mass of }K_2CO_3=\text{ Moles of }K_2CO_3\times \text{ Molar mass of }K_2CO_3

\text{ Mass of }K_2CO_3=(1.0599moles)\times (138.205g/mole)=146.483g

The theoretical yield of potassium carbonate = 146.483 g

The experimental yield of potassium carbonate = 125 g

Now we have to calculate the % yield of potassium carbonate.

Formula for percent yield :

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

\% \text{ yield of }K_2CO_3=\frac{125g}{146.483g}\times 100=85.33\%

Therefore, the % yield of potassium carbonate is, 85.33%

6 0
2 years ago
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