Volume fraction = volume of the element / volume of the alloy
Volume = density * mass
Base: 100 grams of alloy
mass of tin = 15 grams
mass of lead = 85 grams
volume = mass / density
Volume of tin = 15g / 7.29 g/cm^3 = 2.06 cm^3
Volume of lead = 85 g / 11.27 g/cm^3 = 7.54 cm^3
Volume fraction of tin = 2.06 cm^3 / (2.06 cm^3 + 7.54 cm^3) = 0.215
Volume fraction of lead = 7.54 cm^3 / (2.06 cm^3 + 7.54 cm^3) = 0.785
As you can verify the sum of the two volume fractions equals 1: 0.215 + 0.785 = 1.000
Also water H2O is made of H+ and OH- ions. so when an acidic substance is added to water the concentration of H+ ions increase.
<h3><u>Answer;</u></h3>
The value of Kc does not depend on starting concentrations.
<h3><u>Explanation;</u></h3>
- <em><u>At constant temperature, changing the equilibrium concentration does not affect the equilibrium constant, because the rate constants are not affected by the concentration changes. </u></em>
- When the concentration of one of the participants is changed, the concentration of the others vary in such a way as to maintain a constant value for the equilibrium constant.
Answer:
<h2>uranium-235 (²³⁵U) and uranium-238 (²³⁸U)</h2>
Explanation:
The gaseous diffusion process utilizes uranium hexafluoride, UF₆, because although it is a solid at room temperature it is easily vaporized. [1] UF6 is not only convenient for its volatility, but also due to the fact that fluorine only consists of the isotope ¹⁹F, meaning the difference in molecular weights for UF6 are purely reliant on 235U and 238U.Here arises another problem however, for the masses of the two uranium isotopes are so nearly equal there is very little separation of 235UF6 and 238UF6 with one pass through a diffuser.Therefore a cascade process is needed to obtain any measurable amount of enrichment. In a cascade the feed stream at diffuser 1 is the UF6 prior to enrichment (meaning it will contain 0.711% 235U and 99.289% 238U) and marks the start of the cascade. There will be hundreds to thousands of diffusers on the upward or enriching side as well as on the downward or depleted side. The slightly enriched UF6 is sent up the cascade process to the next diffuser where it will be enriched again. The slightly depleted UF6 will be sent downward through the cascade where it will also be enriched again. In this way, the enriched uranium keeps getting enriched and sent onward, and the depleted uranium also gets enriched and sent onward. The depleted uranium always gets sent downward where it will eventually be ejected from the downward stream as depleted uranium.
