Challenge question: This question is worth 6 points. As you saw in problem 9 we can have species bound to a central metal ion. T

Response:

CRYSTAL

A LARGE NUMBER OF ATOMS ORGANIZED IN A REGULAR STRUCTURE

1:1

Reasoning:

To find the answer, start by calculating the total mass of the copper utilized:

Copper used = 100 pennies x 3.0g Cu per penny = 300.0 g Cu

Next, identify the path and molar ratios from Cu produced back to CuFeS2 needed using the established balanced reactions:

1 Cu2S from 2 CuS; 2Cu from 1 Cu2S; 2CuS from 2CuFeS2
Thus, 2Cu comes from 2CuFeS2, indicating a 1:1 molar ratio.

Then convert grams of Cu to moles and grams of CuFeS2:
= 300.0 g Cu * 1 mol Cu/63.546g Cu * 2 mol CuFeS2/2 moles Cu

= 4.72 moles CuFeS2

The required amount of chalcopyrite mined = 4.72 moles CuFeS2 * 183.54 g CuFeS2/1 mole CuFeS2 = 866.49 g CuFeS2

Answer:NH₃/NH₄Cl

The pH of a buffer can be determined using Henderson-Hasselbalch's equation.

When the concentration of acid equals that of the base, the pH aligns with the pKa of the buffer. The ideal pH range is pKa ± 1.

Below are the buffers and their corresponding pKa values:

• CH₃COONa/CH3COOH (pKa = 4.74)

• NH₃/NH₄Cl (pKa = 9.25)

• NaOCl/HOCl (pKa = 7.49)

• NaNO₂/HNO₂ (pKa = 3.35)

• NaCl/HCl Not a buffer

Thus, the ideal buffer is NH₃/NH₄Cl.

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