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Mariana [72]
3 years ago
15

What is the best way to eliminate poison ivy?

Chemistry
1 answer:
antoniya [11.8K]3 years ago
4 0
What poison ivy looks like:
Each leaf has 3 little leaflets. It tends to grow off-white berries on the leaves in the early fall, and grows yellow-green flowers in the spring. 

Treating poison ivy:
If you have a rash from poison ivy, it is recommended to rinse your skin off with lukewarm water immediately after you notice the rash. 
Make sure to wash your clothing since the oil from the plant can stick to your clothing.
Do not scratch at the rash, as scratching will cause an infection.
If blisters appear on your skin, do not mess with them as it can prevent infection.

How to kill poison ivy:
If you find the plant, do not pull or burn it. Even if you pull it, it will still grow out of the ground since you did not eliminate the roots of it. Burning it can release <span>urushiol into the air which can be harmful.
</span>It is recommended to use a specialized weed killing spray. Always look at the ingredients in the spray and look for: Triclopyr, which fights through the surface of the plant, and also <span>glyphosate, that helps kill the roots of the plant.
</span>Once you use a spray, it will take 1-2 weeks for it to die. It will eventually turn yellowish-brownish.

Important:
 Poison ivy contains urushiol which can cause allergic reaction.
Burning poison ivy releases urushiol into the air which can also cause allergic reaction and can be harmful. 


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Explanation:

El óxido de cobre (II), también llamado antiguamente óxido cúprico ({\displaystyle {\ce {CuO}}}{\displaystyle {\ce {CuO}}}), es el óxido de cobre con mayor número de oxidación. Como mineral se conoce como tenorita.

{\displaystyle {\ce {2Cu + O2 = 2CuO}}}{\displaystyle {\ce {2Cu + O2 = 2CuO}}}

Aquí, se forma junto con algo de óxido de cobre (I) como un producto lateral, por lo que es mejor prepararlo por calentamiento de nitrato de cobre (II), hidróxido de cobre (II) o carbonato de cobre (II):

{\displaystyle {\ce {2 Cu(NO3)2 = 2 CuO + 4 NO2+ O2}}}{\displaystyle {\ce {2 Cu(NO3)2 = 2 CuO + 4 NO2+ O2}}}

{\displaystyle {\ce {Cu(OH)2 (s) = CuO (s) + H2O (l)}}}{\displaystyle {\ce {Cu(OH)2 (s) = CuO (s) + H2O (l)}}}

{\displaystyle {\ce {CuCO3 = CuO + CO2}}}{\displaystyle {\ce {CuCO3 = CuO + CO2}}}

El óxido de cobre (II) es un óxido básico, así se disuelve en ácidos minerales tales como el ácido clorhídrico, el ácido sulfúrico o el ácido nítrico para dar las correspondientes sales de cobre (II):

{\displaystyle {\ce {CuO + 2 HNO3 = Cu(NO3)2 + H2O}}}{\displaystyle {\ce {CuO + 2 HNO3 = Cu(NO3)2 + H2O}}}

{\displaystyle {\ce {CuO + 2 HCl =CuCl2 + H2O}}}{\displaystyle {\ce {CuO + 2 HCl =CuCl2 + H2O}}}

{\displaystyle {\ce {CuO + H2SO4 = CuSO4 + H2O}}}{\displaystyle {\ce {CuO + H2SO4 = CuSO4 + H2O}}}

Reacciona con álcali concentrado para formar las correspondientes sales cuprato.

{\displaystyle {\ce {3 XOH + CuO + H2O = X3[Cu(OH)6]}}}{\displaystyle {\ce {3 XOH + CuO + H2O = X3[Cu(OH)6]}}}

Puede reducirse a cobre metálico usando hidrógeno o monóxido de carbono:

{\displaystyle {\ce {CuO + H2 = Cu + H2O}}}{\displaystyle {\ce {CuO + H2 = Cu + H2O}}}

{\displaystyle {\ce {CuO + CO = Cu + CO2}}}{\displaystyle {\ce {CuO + CO = Cu + CO2}}}

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2NO (g) + O2 (g) →2NO2 (g) At equilibrium [NO] = 2.4 × 10 -3 M, [O2] = 1.4 × 10 -4 M, and [NO2] = 0.95 M.
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Answer:

K=1.12x10^9

Explanation:

Hello there!

Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:

K=\frac{[NO_2]^2}{[NO]^2[O_2]}

Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.

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