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Eva8 [605]
3 years ago
9

¿sabes cual es la concentración de azúcar en un refresco de un litro?​

Chemistry
1 answer:
Grace [21]3 years ago
8 0
“Esto quiere decir que en una lata de 330 ml tiene alrededor de 36 gramos de azúcar, mientras que una botella de 1 l tendrá 110 gramos”

110 gramos
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Compounds in alcoholic beverages that enhance flavor and appearance but may contribute to hangover symptoms are called:_____.
olga_2 [115]

Compounds in alcoholic beverages that enhance flavor and appearance but may contribute to hangover symptoms are called congeners.

<h3>Alcoholic beverages:</h3>

Congeners are compounds that add to the flavor, smell, and appearance of most alcoholic beverages. These substances may make hangover symptoms worse. Because they contain fewer congeners than whiskey, brandy, and red wine, clear alcoholic beverages like gin and vodka have less of a hangover-inducing effect.

The impact of ethanol, or the alcohol in your drinks, is the primary contributor to a hangover. It is a poisonous substance that acts as a diuretic in the body, which causes you to urinate more frequently and increases the likelihood that you will become dehydrated. The incidence and intensity of hangovers are both increased by congeners, substances created during the digestion and maturation of alcohol.

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2 years ago
Be sure to answer all parts. A concentration cell consists of two Sn/Sn2+ half-cells. The electrolyte in compartment A is 0.23 M
kolbaska11 [484]

Answer:

Part A. The half-cell B is the cathode and the half-cell A is the anode

Part B. 0.017V

Explanation:

Part A

The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).

So, the half-cell B is the cathode and the half-cell A is the anode.

Part B

By the Nersnt equation:

E°cell = E° - (0.0592/n)*log[anode]/[cathode]

Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.

E°cell = 0 - (0.0592/2)*log(0.23/0.87)

E°cell = 0.017V

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3 years ago
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Explanation:

trust me thats the right answer i am not goo at explanations

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