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3 years ago

What type of bond forms between two oxygen atoms

1 answer:

6 0

Answer:A double convalescent bond is where two pairs of electrons are shared between the atoms rather than just one pair. Two oxygen atoms can both achieve stable structures by sharing two pairs of electrons as in the diagram.

Explanation:

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Write a story of your life when you were hurted by someone whom you trusted blindly...

SOVA2 [1]

Answer:

Sis I think it happened with me but I am not able to remember if u want u can share if it happened with u

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3 years ago

Express in scientific notation. Make sure your answer has the same number of significant figures as the starting value.

frez [133]

The correct response is the second option.
6.1103x10^4. As this was the only answer that had the same number of significant figures as the starting value.

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3 years ago

How many moles of H2 and N2 can be formed by the decomposition of 0.145 mol of ammonia, NH3 ?

zepelin [54]

Answer:

Explanation:

The moles of H2 and N2 are as follows respectively, 0.3915mol of H2 and 0.1305 mol of N2.

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

Which statement correctly identifies the nuclide that is most likely to be unstable and describes why?

iVinArrow [24]

The answer would be B.

U-238 has a n to p ration of 1.6:1. 146 neutrons and 92 protons.

It is actually the most commonly used isotope is reactors.

C-14 is also a radioactive isotope with 8 neutrons and 6 protons.

The usual and ideal n to p ratio is 1:1 such as C-12 or Mg-24

8 0

3 years ago