Answer: The 3rd and 6th bullet point is the quantitative data.
Explanation: Quantitative data is expressed by NUMBERS and Qualitative data is expressed by WORDS. The 3rd and 6th one is correct because they both use numbers to compare how much time hummingbirds spent feeding on nectar.
I am not all understood but for the school to earn money you can:
make
--a raffle
--lotto
-- yard sale
-- class photo
-- origami for sale or something
-- buffet or food sale (example all Friday ice cream sale, 2 livre ice cream)
Answer : The value of 
is 28.97 kJ/mol
Explanation :
To calculate of the reaction, we use clausius claypron equation, which is:
![\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= vapor pressure at temperature 
= 462.7 mmHg
= vapor pressure at temperature 
= 140.5 mmHg
= Enthalpy of vaporization = ?
R = Gas constant = 8.314 J/mol K
= initial temperature = ![-21.0^oC=[-21.0+273]K=252K](https://tex.z-dn.net/?f=-21.0%5EoC%3D%5B-21.0%2B273%5DK%3D252K)
= final temperature = ![45^oC=[-41.0+273]K=232K](https://tex.z-dn.net/?f=45%5EoC%3D%5B-41.0%2B273%5DK%3D232K)
Putting values in above equation, we get:
![\ln(\frac{140.5mmHg}{462.7mmHg})=\frac{\Delta H_{vap}}{8.314J/mol.K}[\frac{1}{252}-\frac{1}{232}]\\\\\Delta H_{vap}=28966.6J/mol=28.97kJ/mol](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7B140.5mmHg%7D%7B462.7mmHg%7D%29%3D%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B252%7D-%5Cfrac%7B1%7D%7B232%7D%5D%5C%5C%5C%5C%5CDelta%20H_%7Bvap%7D%3D28966.6J%2Fmol%3D28.97kJ%2Fmol)
Therefore, the value of is 28.97 kJ/mol
Answer: 35.4 grams
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
where,
Molality = 2.65
n= moles of solute =?
= volume of solution in ml = 445 ml
Putting in the values we get:
Mass of solute in g=
Thus 35.4 grams of 
is needed to prepare 445 ml of a 2.65 m solution of .
