(2 pts) The solubility of InF3 is 4.0 x 10-2 g/100 mL. a) What is the Ksp? Include the chemical equation and Ksp expression. MW

Answer:

yes

Explanation:

yes

We assume that the stated 50% is measured by volume. Molarity defines the concentration in terms of moles of solute per volume of solution.

      To find the moles of NaOH, use: (0.1 moles / L)(0.4 L)
                      n = 0.04 moles of NaOH

Assuming we start with 1 mL of 50% NaOH solution, 
  
                        (1 mL solution)(1.525 g/mL)(0.50) = 0.7625 g
Then, the number of moles calculates as follows,[
                  0.7625 g NaOH x (1 mol / 40 g) = 0.01906 moles of NaOH

                The volume of solution required can be determined by:(0.04 moles of NaOH)(1 mL solution / 0.01906 moles of NaOH)
                 
                 Thus, the needed volume comes out to be 2.09 mL

Answer: 2.09 mL

The quantity is 6.074 X 10¹⁸ molecules. To calculate the molecular weight (MM) of the compound C₂₇H₄₆O, we use the formula: 27*(MM of C) + 46*(MM of H) + (MM of O). This leads to: 27*(12.0107) + 46* (1.00784) + (15.999) = 324.2889 + 46.36064 + 15.999 = 386.64854 g. The molar mass of any compound indicates the number of molecules found in one mole, which is 6.022 X 10²³ (Avogadro's number). So, if 386.64854 g of C₂₇H₄₆O consists of 6.022 X 10²³ cholesterol molecules, we can determine how many molecules are found in a deposit of 3.9mg or 0.0039g of C₂₇H₄₆O by using the unitary method. Number of molecules = 6.074 X 10¹⁸ molecules.

Answer:

C₂H₅O₂

Explanation:

From the information provided in the question, we have the following details:

Mass of compound = 0.25 g

Mass of CO₂ = 0.3664 g

Mass of H₂O = 0.15 g

Empirical formula =?

Now, let’s calculate the masses of carbon, hydrogen, and oxygen within the compound as follows:

For Carbon (C):

Mass of CO₂ = 0.3664 g

Molar mass of CO₂ = 12 + (2×16) = 44 g/mol

Mass of C = 12/44 × 0.3664

Mass of C = 0.1

For Hydrogen (H):

Mass of H₂O = 0.15 g

Molar mass of H₂O = (2×1) + 16 = 18 g/mol

Mass of H = 2/18 × 0.15

Mass of H = 0.02 g

For Oxygen (O):

Mass of C = 0.1 g

Mass of H = 0.02 g

Mass of compound = 0.25 g

Mass of O =?

Mass of O = (Mass of compound) – (Mass of C + Mass of H)

Mass of O = 0.25 – (0.1 + 0.02)

Mass of O = 0.25 –0.12

Mass of O = 0.13 g

In conclusion, we will now find the empirical formula for the compound:

C = 0.1

H = 0.02

O = 0.13

Next, we divide by their respective molar mass:

C = 0.1 / 12 = 0.0083

H = 0.02 / 1 = 0.02

O = 0.13 / 16 = 0.0081

Then we divide by the smallest value:

C = 0.0083 / 0.0081 = 1

H = 0.02 / 0.0081 = 2.47

O = 0.008 / 0.008 = 1

Finally, we multiply by 2 to present in whole numbers:

C = 1 × 2 = 2

H = 2.47 × 2 = 5

O = 1 × 2 = 2

Therefore, the empirical formula for the compound is C₂H₅O₂

D.) Gas X possesses a lower density and effuses more quickly than Gas Y. Explanation: Gas X has a lower molar mass compared to Gas Y, and the density of a gas is directly proportional to its molar mass. Therefore, it has a lower density than Gas Y. Thomas Graham discovered that, under constant temperature and pressure, the effusion rates of various gases are inversely related to the square root of their molar masses (M). Consequently, a gas with a lower molar mass will effuse faster than one with a higher molar mass. Thus, Gas X effuses faster than Gas Y. So, the correct selection is: D.) Gas X has a lower density and effuses faster than Gas Y.