An impure sample of zinc (zn) is treated with an excess of sulfuric acid (h 2 so 4) to form zinc sulfate (znso4) and molecular h

Answer:

The concentration of P in the pond at equilibrium is 0.034 g/m³

Explanation:

Given the total mass = 49.9 g

1 day = 24 hours

mass per hour;

Incoming mass = (49.9 g / day) * (1 day /24 hr )

            = 2.079 g/hr

Outgoing mass = 0

Mass lost due to sunlight = k V  

Given the half-life = 3.4 hours

For a first-order reaction; k, the rate constant = ln2/t, where t is the half-time

                     ln 2= 0.693, V= volume

                     k = 0.693 / t_half = 0.693 / 3.4 = 0.2038 hr⁻¹

Substituting all parameters into the equation k V;

Mass lost to sunlight = k V  

= Incoming mass per hour / kV

= 2.079 g/hr / (0.2038 hr⁻¹ x 300 m³)

=  

0.034 g/m³

Answer:

The amount of calcium sulfate that precipitates is 6.14 grams.

Explanation:

Step 1: Provided data

We are mixing 500.0 mL of 0.10 M Ca^2+ with 500.0 mL of 0.10 M SO4^2−

The Ksp for CaSO4 is 2.40×10^−5.

Step 2: Determine moles of Ca^2+

Moles of Ca^2+ = Molarity of Ca^2+ * Volume

Moles of Ca^2+ = 0.10 * 0.500 L

Moles of Ca^2+ = 0.05 moles

Step 3: Determine moles of SO4^2-

Moles of SO4^2- = 0.10 * 0.500 L

Moles SO4^2- = 0.05 moles

Step 4: Compute total volume

Total volume = 500.0 mL + 500.0 mL = 1000 mL = 1L

Step 5: Compute Q

Q = [Ca2+] [SO42-]

[Ca2+] = 0.050 M and [SO42-]

Qsp = (0.050)(0.050) = 0.0025 >> Ksp

This indicates that precipitation will take place.

Step 6: Calculate molar solubility

Ksp = 2.40 * 10^-5 = [Ca2+][SO42-] =(x)(x)

2.40 * 10^-5 = x²

x = √(2.40 * 10^-5)

x = 0.0049 M (molar solubility)

Step 7: Determine total dissolved CaSO4

Total CaSO4 dissolved = 0.0049 M * 1 L * 136.14 g/mol = 0.667 g

Step 8: Calculate initial mass of CaSO4

Initial moles of CaSO4 = 0.050

Initial mass of CaSO4 = 0.050 * 136.14 g/mol

Initial mass of CaSO4 = 6.807 grams

Step 9: Calculate precipitate mass

6.807 - 0.667 = 6.14 grams.

The mass of calcium sulfate that will emerge as a precipitate is 6.14 grams.

First, we need to identify the half-reaction for magnesium. It can be represented as:

Mg2+ + 2e- = Mg

Next, we will determine the overall charge generated during the electrolysis using the information derived from the half-reaction. The calculation follows:

4.50 kg Mg (1000 g / 1 kg) (1 mol / 24.305 g) (2 mol e- / 1 mol Mg) (96500 C / 1 mol e-) = 35733388.2 C

The provided EMF is given in voltage. Since 1 V equals J/C, 5 V translates to 5 J/C.

Therefore, 35733388.2 C (5 J/C) = 178666941 J
Finally, 178666941 J (1 kW-h / 3.6x10^6 J) = 49.63 kW-h

The maximum wavelength of light required to break the amide bond is 268 nm. First, we find the average bond energy, then use Avogadro's number to figure out the energy needed for one molecule. Finally, applying the relationship between energy and wavelength, we can conclude the value of the wavelength.

3.677 mg =