Hello,
Here is your answer:
The proper answer to this question is option C "stigma".
Here is how:
The stigma is responsible for producing pollen in a plant.
Your answer is C.
If you need anymore help feel free to ask me!
Hope this helps!
First, consider the steps to heat the sample from 209 K to 367K.
1) Heating in liquid state from 209 K to 239.82 K
2) Vaporaizing at 239.82 K
3) Heating in gaseous state from 239.82 K to 367 K.
Second, calculate the amount of heat required for each step.
1) Liquid heating
Ammonia = NH3 => molar mass = 14.0 g/mol + 3*1g/mol = 17g/mol
=> number of moles = 12.62 g / 17 g/mol = 0.742 mol
Heat1 = #moles * heat capacity * ΔT
Heat1 = 0.742 mol * 80.8 J/mol*K * (239.82K - 209K) = 1,847.77 J
2) Vaporization
Heat2 = # moles * H vap
Heat2 = 0.742 mol * 23.33 kJ/mol = 17.31 kJ = 17310 J
3) Vapor heating
Heat3 = #moles * heat capacity * ΔT
Heat3 = 0.742 mol * 35.06 J / (mol*K) * (367K - 239.82K) = 3,308.53 J
Third, add up the heats for every steps:
Total heat = 1,847.77 J + 17,310 J + 3,308.53 J = 22,466.3 J
Fourth, divide the total heat by the heat rate:
Time = 22,466.3 J / (6000.0 J/min) = 3.7 min
Answer: 3.7 min
Answer:
oxygen molecule
Explanation:
I belive it is oxygen molecule
Answer:
Volume occupied by oxygen gas at 15 degree centigrade is equal to
centimeter cube
Explanation:
Assuming Pressure is constant.

where T1 and T2 are temperature in Kelvin
Substituting the give values we get-


Volume occupied by oxygen gas at 15 degree centigrade is equal to
centimeter cube
By dividing the percentage composition with the molar mass of that element we will get the empirical formula. Then using that empirical formula and formula mass we can find the molecular formula.
<u>Explanation:</u>
The chemical properties of any substance are defined obviously by the different types and relative amounts of atoms constituting its primary entities (in case of covalent compounds the primary entities are molecules and ions in the event of ionic compounds).
A percent composition of any compound gives the mass percent of each element present in the compound; in addition to that frequently it is determined experimentally and utilized to derive an empirical formula of any compound. An empirical formula mass of any covalent compound could be comparable with the molar or molecular mass of a compound to acquire a molecular formula.