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3 years ago

6. What is the similarity between an oxbow lake and a delta?

1 answer:

3 0

The similarity between an oxbow lake and delta is that both are formed by the deposition of sediments. Sediments deposit between the path of river, so forms a curve structure called oxbow lake.

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Cyanide and water react in a proton transfer reaction to form hydrogen cyanide and hydroxide

alukav5142 [94]

The proton transfer reaction between Cyanide and water can be written as; X^- + H2O -----> HX + OH^-

<h3>What is a proton transfer reaction?</h3>

A proton transfer reaction is one in which a proton is moved from one chemical specie to another.It is in fact and acid - base reaction in the Brownstead - Lowry sense.

The proton transfer reaction between Cyanide and water can be written as(Let the cyanide ion be shown as X);

X^- + H2O -----> HX + OH^-

Learn more about proton transfer: brainly.com/question/861100?

4 0

2 years ago

Read 2 more answers

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

What is an example of entropy from everyday life?

Deffense [45]

A Campfire is an example, the solid wood becomes ash.

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3 years ago

State two procedures which should be used to properly handle a light microscope

babunello [35]

hold at base and arm

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3 years ago

What is the number after an element in a chemical equation called? Example: H2

Sophie [7]

Answer:

the answer to your problem is A.Subscript...BRAINLIEST

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3 years ago