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evablogger [386]
3 years ago
7

Which landforms are created where cirques meet?

Chemistry
1 answer:
dezoksy [38]3 years ago
6 0

Aretes and horns

Explanation:

Aretes and horns are the landforms that are created where cirques meet. These landforms are products of glacier erosion in most temperate and polar regions.

  • Cirques are common products of alpine glacier erosion.
  • They arise in form of bowl shaped hollows on the side of a mountain.
  • These hollows joins to form a U-shaped valley called Aretes and unique pointed arcs known as horns.

Learn more:

factors that affects erosion brainly.com/question/2473244

#learnwithBrainly

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C. Sb has a lower ionization energy and a lower electronegativity than I.

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The chemical elements present higher ionization energy the higher and to the right of the periodic table they are.

Electronegativity usually increases from left to right over a period of the periodic table and from top to bottom in the groups.

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Which of the following is an important application of hydrogen bonding
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I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r
maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P

Now, the moles of Li3P consumed by 15 g of Al2O3:

n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP

Therefore, the total mass of products is:

m_{products}=13g+8.5g\\\\m_{products}=21.5g

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

7 0
2 years ago
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