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:
Rubidium= [Kr] 5s^1
Calcium= [Ar] 4s^2
Aluminium= [Ne] 3s^2 3p^1
Explanation:
A noble gas configuration begins with the elemental symbol of the last noble gas prior to the atom. The symbol is then followed by the remaining electrons.
Hope this helped! good luck :)
A Nonrewnewable source is hard to get back from its source, while a renewable source is easy to get back from its source.
Example:
Nonrenewable: crude oil, oil, coal, fossil fuels, etc.
Renewable: sunlight, wind, biomass, rain, tides, waves and geothermal heat, etc.
Answer:

Explanation:
Hello,
In this case, by using the ideal gas equation, we first compute the moles of oxygen at the given volume, pressure and temperature:

Then, since molar mass of gaseous oxygen is 32 g/mol, we compute the contained mass in grams as shown below:

Best regards.