The amount of alcohol contained in each of the listed beverages can be compared as follows; 1/2 oz of 80 proof liquor > 5 oz of wine (12% alcohol) > 12 oz of wine cooler (5%alcohol) = 12 oz of beer (5%alcohol) .
A drink that contains alcohol must be an intoxicating beverage. The extent of intoxication of an alcoholic beverage depends on the amount of alcohol that the beverage contains. There are various types of alcoholic beverages as listed in the question.
The amount of alcohol contained in each of the listed beverages can be compared as follows; 1/2 oz of 80 proof liquor > 5 oz of wine (12% alcohol) > 12 oz of wine cooler (5%alcohol) = 12 oz of beer (5%alcohol) .
Learn more about alcoholic beverage: brainly.com/question/6967136
The answer would be color. The temperature affects the brightness and the shade of the star.
Answer:
There is a lot of biodiversity in this seafloor.
Explanation:
Biodiversity is just the diversity in species and animals in a certain environment.
Answer:
1.09 x 10⁻⁴ M
Explanation:
The equation of the reaction in given by
Ni²⁺ (aq) + 6NH₃ (aq) ⇔ Ni(NH₃)₆
At the beginning o the reaction, we have 0.18M concentration of Ni and 1.2M concentration of aqueous NH₃ and zero concentration of the product
As the reaction proceeds towards equilibrium, the concentration of the reactants decrease as the concentration of the product starts to increase
From the equation,
1 mole of Ni²⁺ reacts with 6 moles of aqueous NH₃ to give 1mole of Ni(NH3)
therefore
0.18 M of Ni would react with 1.08M (6 x 0.18M) aqueous NH₃ to give 0.18M of Ni(NH₃)6
At equilibrium,
1.08M of NH3 would have reacted to form the product leaving
(1.2 - 1.08)M = 0.12M of aqueous NH₃ left as reactant.
Therefore, formation constant K which is the ratio of the concentration of the product to that of the reactant is given by
K = [Ni(NH₃)₆} / [Ni²⁺] 6[NH₃]
5.5 x 10⁸ = 0.18 M / [Ni²⁺] [0.12]⁶
[Ni²⁺]= 0.18 M / (5.5 x 10⁸) (2.986 x 10⁻⁶)
=0.18 M / 0.00001642
= 1.09 x 10⁻⁴ M
[Ni]²⁺ = 1.09 x 10⁻⁴ M
Hence the concentration of Ni²⁺ is 1.09 x 10⁻⁴ M
