what’s the empirical and molecular formula for a compound with 20.2% sodium, 37.6% sulfur, 42.2%oxygen and a molar mass of 682.8

Answer:

B.  26.0 μL.

Explanation:

Hello,

Considering the provided mass and density, the volume calculates to be:

Thus, the solution is B.  26.0 μL.

Best regards.

Answer:

45727g

Explanation:

The overall ionic reaction can be described as follows:

CrO42^- + 3 Fe2^+ + 8 H2O ------> Cr(OH)^3(s) + 3 Fe(OH)^3(s) + 4 H^+.

The amount of wastewater processed each day is specified as 60m^3/h, with the concentration of chromium in the wastewater recorded at 4.0 mg/L, and the allowable discharge concentration is set at 0.1 mg/L.

Step one: Convert m^3/h to L/h. Therefore, 60 m^3/h × 1000 dm^3 = 60000 L/h.

Step two: Calculate the amount of chromium consumed.

The amount of chromium utilized = { 60,000 × ( 4.0 - 0.1) } ÷ 1000 = 234 g.

Step three: Calculate the masses of Cr(OH)3 and Fe(OH)3.

The moles of chromium = 234/52 = 4.5 moles.

Molar mass of Cr(OH)3 = 103 g/mol and the molar mass of Fe(OH)3 = 106.8 g/mol.

Thus, the mass of Cr(OH)3 = 4.5 × 103 = 463.5 g.

And, the mass of Fe(OH)3 = 13.5 × 106.8 = 1441.8 g.

Consequently, the total equals 463.5 g + 1441.8 g = 1905.3 g.

Step four: Calculate the amount of particulate matter generated each day.

The total particulate matter produced daily = 24 × 1905.3 = 45727g.

Response:

a. To purify a gypsum sample, you will need the following equipment: Bunsen burner, beaker, filter funnel, stirring rod, and filter paper.

b. Gypsum, a sulfate mineral consisting of calcium sulfate dihydrate, can be purified by following these steps:

1. Add water to the gypsum in a beaker.

2. Stir the mixture thoroughly with the stirring rod.

3. Use the filter paper and filter funnel to remove excess solids from the mixture.

4. Heat the filtered mixture on the Bunsen burner to evaporate the remaining water.

5. After cooling, filter again through the filter paper to obtain pure gypsum.

Answer: The number of sulfur dioxide molecules present is 1.27·10²³.
Calculating: m(SO₂) equals 13.5 g.
Using the formula n(SO₂) = m(SO₂) ÷ M(SO₂).
This gives n(SO₂) = 13.5 g ÷ 64 g/mol.
Resulting in n(SO₂) = 0.21 mol.
Subsequently, N(SO₂) = n(SO₂) ·Na.
Therefore, N(SO₂) = 0.21 mol · 6.022·10²³ 1/mol.
Ultimately, N(SO₂) equals 1.27·10²³.
Where n represents amount of substance.
M refers to molar mass.
Na is Avogadro's number.

The enthalpy change associated with the precipitation reaction is 84 kJ/mole

Why?

The chemical equation for the reaction can be written as

AgNO₃(aq) + NaCl (aq) → AgCl(s) + NaNO₃(aq)

To determine the enthalpy change, the following equation applies

To calculate the heat (Q):

Next, we need to calculate the number of moles involved in the reaction (n):

With these two values, we can substitute them into the first equation:

Have a great day!