Response:
of 
can be found in 39.5 grams of .
Clarification:
Atomic weights: P= 31, F= 19,
The molar mass equals 1 atomic weight of P + 5 atomic weights of
F= 31+5 × 19
= 31+95
=126 g/mole
The number of moles in 39.5 gm of
equals 
= 
=0.3134 moles
1 mole of any substance encompasses
0.3131 moles comprises 0.3134
Thus, of can be found in 39.5 grams of .
To determine the answer, you need to understand the formula for converting grams to moles, which will then lead you to the number of molecules.
The result is 2 moles of N2O5. The process is as follows:
(0.25 g N2O5) (1 mol/ 108 g)=2.31 molecules
Thus, the final answer is 2 molecules.
Answer:
The glycerol solution has a molality of 2.960×10^-2 mol/kg.
Explanation:
Calculating the moles of glycerol involves the formula: Moles = Molarity × Volume of solution = 2.950×10^-2 M × 1 L = 2.950×10^-2 moles.
To find the mass of water, use: Mass = Density × Volume = 0.9982 g/mL × 998.7 mL = 996.90 g, which converts to 0.9969 kg.
The formula for molality is: Molality = Moles of solute/Mass of solvent (in kg) = 2.950×10^-2/0.9969 = 2.960×10^-2 mol/kg.
This procedure entails diluting the 12 molar HCl. To decrease the concentration, we must create an equation to determine how much of the 12M is needed for the 3.5M solution.
12 moles HCl 3.5 moles HCl
——————— = ———————
1 Liter of Soln ‘x’ Liters of Soln
Note that the ratio of 12 moles over 1 liter corresponds to 12 molar; thus, we maintain the original concentration of the HCl. By equating it to the 3.5 over ‘x’, we are still preserving the concentration.
After computation, we determine ‘x’ to be 0.292. This value indicates that within 0.292 liters of our 12 M HCl solution, there are 3.5 moles of HCl. Yet, we are not finished.
0.292 liters of 12 M HCl can create 1 liter of 3.5 M HCl, but the inquiry demands 1.5 liters. To achieve this, multiply 0.292 liters by 1.5, resulting in 0.4375, which denotes the quantity of 12 M HCl necessary to prepare a 1500 mL 3.5 M HCl solution.
