Given the connection between Aw and K (Aw=2k) could you use the ideal gas law and derive the Boltzmann constant. Water freezes a

Answer:

Explanation:

We have:

V₁ = 18.5 L

T₁ = 18.5° C = 273 + 18.5 = 291.5 K

V₂ = 19.8 L

T₂ =?

Pressure remains constant

Applying the ideal gas law

Answer:

Empirical formula: Na3S4O9

Molecular formula: Na6S8O18

Explanation:

1) Upang makuha ang empirical formula, na siyang pinaka-simpleng buong bilang na ratio ng bawat elemento sa hindi kilalang compound na ito, una nating hahatiin ang dami ng bawat elemento sa kanilang sariling atomic mass.

Na= 23g/mol, S= 32g/mol, O=16g/mol

Na = 20.2/23 = 0.878mol

S = 37.6/32 = 1.175mol

O = 42.2/16 = 2.638mol

Susundan ito sa pamamagitan ng paghahati ng bawat bilang ng moles sa pinakamababang halaga (0.878mol)

Na = 0.878/0.878 = 1

S = 1.175/0.878 = 1.34

O = 2.638/0.878 = 3.004

Susunod, paramihin ang bawat decimal na bahagi ng 3 upang maging buo ang mga numero:

Na = 3

S= 4.02 ~ 4

O= 9.012 ~ 9

Kaya, ang pinaka-simpleng ratio ng Na, S, at O ay 3:4:9 kaya, ang empirical formula ay Na3S4O9

2) Para makuha ang molecular formula, gagamitin natin ang molecular weight ng bawat elemento sa empirical ratio, at ang kabuuang molar mass ng compound i.e.

Molar mass ng compound= 682.8g/mol

molecular weight ng Na3S4O9 = (3 x 23g/mol) + (4 x 32 g/mol) + (9 x 16 g/mol)

= 69 + 128 + 144

= 341 g/mol

Kaya, (Na3S4O9)n = 682.8g/mol

= (341)n = 682.8

n = 682.8/341

n~ 2

Kaya, molecular formula = (Na3S4O9)2

= Na6S8O18

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

Correct answer:

C) Ca(OH)2

Note: The listed options in the question are incorrect. The accurate selections are shown below:

15. A total of 3.705 kg of substance Y is required to neutralize 100 moles of HCl(aq).

What might substance Y be? A Ca B CaO C Ca(OH)2 D CaCO3

Explanation:

A) Ca + 2HCl ---> CaCl2 + H2

Molar mass of Ca = 40 g/mol

The mass of Ca needed to neutralize 100 moles of HCl = 1/2 * 100 * 40 g = 2000g or 2 kg

B) CaO + 2HCl ---> CaCl2 + H2O

Molar mass of CaO = 56.1 g/mol

The mass of CaO needed to neutralize 100 moles of HCl = 1/2 * 100 * 56.1 g = 2805 g or 2.805 kg

C) Ca(OH)2 + 2HCl ---> CaCl2 + 2H2O

Molar mass of Ca(OH)2 = 74.1 g/mol

The mass of Ca(OH)2 necessary to neutralize 100 moles of HCl = 1/2 * 100 * 74.1 g = 3705 g or 3.705 kg

D) CaCO3 + 2HCl ---> CaCl2 + H2O + CO2

Molar mass of CaCO3 = 100 g/mol

The mass of CaCO3 required to neutralize 100 moles of HCl = 1/2 * 100 * 100 = 5000 g or 5.00 kg

Thus, the right choice is C.