Calculate the gravimetric factor for Ag2O in AgS.

The energy released results in a kinetic energy of 92.2 keV for the products. We should convert keV into Joules, noting that 1 keV equals a kiloelectron volt. The required conversion is: 1.602×10⁻¹⁹ <span>joule = 1 eV

Kinetic energy = 92.2 keV * (1,000 eV/1 keV) * (</span>1.602×10⁻¹⁹ joule/1 eV) = 5.76×10²³ Joules

Next, we can determine the velocity of each He atom from the kinetic energy:
KE = 1/2*mv²
5.76×10²³ Joules = 1/2*(4)(v²)
This solves to give us: v = 5.367×10¹¹ m/s

Explanation:

5.5 billion grains of sand (5.5×10^9 grains)

Assuming the amount of brown sand is 6.0%, how many brown grains are found in the bucket?

Grains of brown sand = Percentage of brown sand * Total sand in the bucket

Calculated brown grains = 0.06 * 5.5×10^9 = 0.33 x 10^9 = 3.3 x 10^8 grains

If the concentration of brown sand is 6.0 ppm, what is the number of brown grains in the bucket?

Grains of brown sand = Concentration of brown sand * Total sand in the bucket

6 ppm = 6 / 1,000,000 = 0.000006

Calculated brown grains = 0.000006 * 5.5×10^9 = 3.3 x 10^4 grains

If the concentration of brown sand is 6.0 ppb, how many brown grains exist in the bucket?

Grains of brown sand = Concentration of brown sand * Total sand in the bucket

6 ppb = 6 / 1,000,000,000 = 0.000000006

Calculated brown grains = 0.000000006 * 5.5×10^9 = 3.3 x 10^1 = 33 grains

Answer: 0.0164 molar concentration of hydrochloric acid in the resulting solution.

Explanation:

1) Molarity of 0.250 L HCl solution: 0.0328 M

The amount of HCl in the 0.250 L solution = 0.0082 moles

2) Molarity of 0.100 L NaOH solution: 0.0245 M

The amount of NaOH in the 0.100 L solution = 0.00245 moles

3) Determining the concentration of hydrochloric acid in the final solution.

0.00245 moles of NaOH will neutralize 0.00245 moles of HCl from the original 0.0082 moles of HCl.

The total volume of the mixture becomes 0.100 L + 0.250 L = 0.350 L

Remaining moles of unreacted HCl = 0.0082 moles - 0.00245 moles = 0.00575 moles

Concentration of the remaining HCl:

0.0164 molar concentration of hydrochloric acid in the resulting solution.

Answer: The change in enthalpy for each mole of zinc involved in the reaction is 152.4 kJ/mol.

Explanation:

First, we need to determine the moles of Zn and HCl.

The molar mass of Zn is 65 g/mole

and,

Next, we must identify the limiting reagent and the excess reagent.

The chemical reaction given is:

According to the balanced reaction we find that

1 mole of reacts with 2 moles of

Thus, 0.0206 moles of react with moles of

This leads us to determine that is the excess reagent because the moles provided exceed the required moles, while is limiting and restricts product formation.

Now to find the enthalpy change for each mole of zinc reacting in this reaction.

From the reaction we gather that,[ [TAG_59]]

0.0206 moles of Zn yield heat = 3.14 kJ

This implies that 1 mole of Zn generates heat =

Hence, the enthalpy change per mole of zinc involved in this reaction amounts to 152.4 kJ/mol.