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.
Learn more about congeners here:
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Answer:
2.387 mol/L
Explanation:
The reaction that takes place is:
- 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O
First we <u>calculate how many moles of each reagent were added</u>:
- HCl ⇒ 200.0 mL * 3.85 M = 203.85 mmol HCl
- Ba(OH)₂ ⇒ 100.0 mL * 4.6 M = 460 mmol Ba(OH)₂
460 mmol of Ba(OH)₂ would react completely with (2*460) 920 mmol of HCl. There are not as many mmoles of HCl so Ba(OH)₂ will remain in excess.
Now we <u>calculate how many moles of Ba(OH)₂ reacted</u>, by c<em>onverting the total number of HCl moles to Ba(OH)₂ moles</em>:
- 203.85 mmol HCl *
= 101.925 mmol Ba(OH)₂
This means the remaining Ba(OH)₂ is:
- 460 mmol - 101.925 mmol = 358.075 mmoles Ba(OH)₂
There are two OH⁻ moles per Ba(OH)₂ mol:
- OH⁻ moles = 2 * 358.075 = 716.15 mmol OH⁻
Finally we <u>divide the number of OH⁻ moles by the </u><u><em>total</em></u><u> volume</u> (100 mL + 200 mL):
- 716.15 mmol OH⁻ / 300.0 mL = 2.387 M
So the answer is 2.387 mol/L
I think the answer is number (4)
<span> Ag(NH3)2Cl + 3HNO3 = AgNO3 +2NH4NO3 + HCl </span>
<span>or
Ag(NH3)2Cl + HNO3 = Ag(NH3)2NO3 + HCl this the complete balanced equation
now remove spectator ions to get net ionic equation
so
</span>
<span>
2H+ + 2NO3- + [Ag(NH3)2]+ Cl- -> AgCl + 2NH4+ + 2NO3- 2NO3- 2H+ [Ag(NH3)2]+ + Cl- -> AgCl + 2NH4+
</span>hope it helps