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V125BC [204]
3 years ago
5

Oil coats the surface of the water, endangering sea life during an oil spill. Why doesn’t oil dissolve in ocean water?

Chemistry
1 answer:
masya89 [10]3 years ago
6 0
The best and most correct answer among the choices provided by your question is the fourth choice or letter D "<span>Oil is nonpolar, and water is polar."

</span>Oil<span> will not </span>dissolve<span> in </span>water<span> because </span>water<span> forms a polar covalent bond and </span>oil<span> forms a non-covalent bond. According to the U.S. Department of Energy, in chemistry molecules </span>dissolve<span> with other molecules that share similar characteristics.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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Endothermic or exothermic
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2 years ago
) Given the following balanced equation, determine the rate of reaction with respect to [O2]. If the rate of formation of O2is 7
Travka [436]

Answer:

Rate of the reaction is 0.2593 M/s

-0.5186 M/s is the rate of the loss of ozone.

Explanation:

The rate of the reaction is defined as change in any one of the concentration of reactant or product per unit time.

2O_3\rightleftharpoons 3O_2

Rate of formation of oxygen : 7.78\times 10^{-1} M/s

Rate of the reaction(R) =\frac{-1}{2}\frac{d[O_3]}{dt}=\frac{1}{3}\frac{d[O_2]}{dt}

R=\frac{1}{3}\frac{d[O_2]}{dt}

Rate of formation of oxygen=3 × (R)

7.78\times 10^{-1} M/s=3\times R

Rate of the reaction(R): 0.2593 M/s

Rate of the reaction is 0.2593 M/s

Rate of disappearance of the ozone:

R=-\frac{1}{2}\frac{d[O_3]}{dt}

\frac{d[O_3]}{dt}=-2\times R=-2\times 0.2593\times M/s=-0.5186M/s

-0.5186 M/s is the rate of the loss of ozone.

6 0
3 years ago
Consider the balanced chemical equation for the combustion of methane (CH4).
Mrac [35]
<h3>Answer:</h3>

                  89.6 L of O₂

<h3>Solution:</h3>

The balanced chemical equation is as,

                                CH₄  +  2 O₂    →    CO₂  +  2 H₂O

As at STP, one mole of any gas (Ideal gas) occupies exactly 22.4 L of Volume. Therefore, According to equation,

             44 g ( 1 mol) CO₂ is produced by  =  44.8 L (2 mol) of O₂

So,

                  88 g CO₂ will be produced by  = X L of O₂

Solving for X,

                        X = (88 g × 44.8 L) ÷ 44 g

                        X = 89.6 L of O₂

5 0
3 years ago
Read 2 more answers
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The answer is 8 :)
All nobel gases have 8 outer electrons.
6 0
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