Answer:
Saturated solution = 180 gram
Explanation:
Given:
Solubility of Z = 60 g / 100 g water
Given temperature = 20°C
Amount of water = 300 grams
Find:
Saturated solution
Computation:
Saturated solution = [Solubility of Z] × Amount of water
Saturated solution = [60 g / 100 g] × 300 grams
Saturated solution = [0.6] × 300 grams
Saturated solution = 180 gram
Answer:
A. False.
Every substance contains the same number of molecules i.e 6.02x10^23 molecules
B. False.
Mass conc. = number mole x molar Mass
Mass conc. of 1mole of N2 = 1 x 28 = 28g
Mass conc. of 1mol of Ar = 1 x 40 = 40g
The mass of 1mole of Ar is greater than the mass of 1mole of N2
C. False.
Molar Mass of N2 = 2x14 = 28g/mol
Molar Mass of Ar = 40g/mol
The molar mass of Ar is greater than that of N2.
Explanation:
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:
brainly.com/question/1837839
#SPJ4
Answer:
<h2>0.93 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
From the question
mass = 37.2 g
volume = 40 mL
We have
We have the final answer as
<h3>0.93 g/mL</h3>
Hope this helps you
Combustion equation of n-hexane:
2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O
a)
Assuming we have 100 moles of air,
Oxygen = 20.9 moles
n-hexane required = 20.9/19 x 2
= 2.2 moles
LFL = Half of stoichometric amount = 2.2 / 2 = 1.1
LFL n-hexane = 1.1%
b)
1.1 volume percent required for LFL
1.1% x 1
= 0.0011 m³ of n-hexane required
