Answer:
Cost to supply enough vanillin is 
Explanation:
Threshold limit of vanillin in air is
per litre means there should be
of vanillin in 1L of air to detect aroma of vanillin.

So, 
So amount of vanillin should be present to detect = 
As cost of 50 g vanillin is
therefore cost of
vanillin = 
Answer:
2.5 g/cm3 .
Explanation:
Density is mass in grams over volume in cubic centimeters. So it is 25 g 10 cm 3 =2.5 g/cm 3 The unit is grams per cubic centimeter. NB - density can also be kilograms over cubic meters Hope this helps!
We can solve the equation and show the solution below:
Oxygen atomic number is 16.
Phosphorus atomic number is 32.
We have the molecular weight:
Molecular weight = (31*4) + (16*10)
Molecular weight = 284 grams/mol
Solving for the grams:
0.4 mole (for P4) * (1 mol P4O10/1 mol P4) * (284 grams P4O10/1 mole P4O10)
Total grams = 113.6
The answer is 113.6 grams.
Answer:
Assuming pressure is held constant,the question reduces to a ratio and proportion type of question where;
At constant pressure,
21.0ft³-55.0°F
11.0fy³=11.0ft³/21.0ft³×55°F
The temperature is 28.809°F≈29°F