Answer:
On the picture are all principles cited on wikipedia.
Explanation:
So the answer is
1. Environment must be exploited to improve living standards
2. Flourishing human and nonhuman life depends on diversity of life forms
Answer:
Molar concentration of the weak acid solution is 0.0932
Explanation:
Using the formula: 
Where Ca = molarity of acid
Cb = molarity of base = 0.0981 M
Va = volume of acid = 25.0 mL
Vb = volume of base = 23.74 mL
na = mole of acid
nb = mole of base
Since the acid is monopromatic, 1 mole of the acid will require 1 mole of NaOH. Hence, na = nb = 1
Therefore, 
Ca = 0.0981 x 23.74/25.0
= 0.093155 M
To 4 significant figure = 0.0932 M
Answer:
The freezing point will be 
Explanation:
The depression in freezing point is a colligative property.
It is related to molality as:

Where
Kf= 
the molality is calculated as:




Depression in freezing point = 
The new freezing point = 
Answer:
- Molar mass = 608.36 g/mol
Explanation:
It seems the question is incomplete. However a web search us shows this data:
" Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner R. B. Woodward. It is used as a tranquilizer and sedative. When 1.00 g reserpine is dissolved in 25.0 g camphor, the freezing-point depression is 2.63 °C (Kf for camphor is 40 °C·kg/mol). Calculate the molality of the solution and the molar mass of reserpine. "
The <em>freezing-point depression</em> is expressed by:
We put the data given by the problem and <u>solve for m</u>:
- 2.63 °C = 40°C·kg/mol * m
For the calculation of the molar mass:<em> Molality</em> is defined as moles of solute per kilogram of solvent:
- 0.06575 m = Moles reserpine / kg camphor
- 25.0 g camphor ⇒ 25.0/1000 = 0.025 kg camphor
We<u> calculate moles of reserpine:</u>
- 0.06575 m = Moles reserpine / 0.025 kg camphor
- Moles reserpine = 1.64x10⁻³ mol
Finally we use the mass of reserpine and the moles to calculate <u>the molar mass</u>:
- 1.00 g reserpine / 1.64x10⁻³ mol = 608.36 g/mol
<em>Keep in mind that if the data in your problem is different, the results will be different. But the solving method remains the same.</em>