True
After the process of Ore processing, Enrichment, Fuel production and being passed through the reactor core the last remaining step is spent fuel disposal.
Answer:
12 atm
Explanation:
First, let us convert Celcius into Kelvin: 28.0 °C = 301.15 K and 129.0 °C = 402.15 K
For this question we must employ the Combined Gas Law: 
, where 
is the initial pressure and 
is the new pressure.
We know that intitially, P=9 atm, V=30 L, and T=301.15K. From our problem, only temperature and pressure changes, while the number of moles, volume and the gas constant, R, stay the same, so they are irrelevant.
Thus, the filled out Combined Gas Law would be:
=
, where the volume, moles of gas, and R are cancelled out.
We can manipulate this equation to derive the new pressure. We find that
9atm≈0.74885.
This means that
≈9/0.74885≈12 atm
This question requires the knowledge of density.
The density of ethyl alcohol = 789 kg m⁻³
The density of water = 1000 kg m⁻³
Density = Mass / Volume
By applying ethyl alcohol,
789 kg m⁻³ = Mass / 0.9 m³
Mass = 710.1 kg
hence the mass of 0.9 m³ ethyl alcohol is 710.1 kg.
Then by applying water,
1000 kg m⁻³ = 710.1 kg / Volume
Volume = 0.7101 m³
= 0.7 m³
hence the equal water volume is 0.7 m³
The solubility equilibrium of 
:
[tex] CaCrO_{4}(aq)<===>Ca^{2+}(aq) + CrO_{4}^{2-}(aq)\\
Q_{sp}=[Ca^{2+}][CrO_{4}^{2-}]\\
= (0.0200 M)(0.0300 M) \\
= 0.0006
Ksp (0.00071) > Qsp (0.0006). So, <u>no precipitate would form</u>.
Answer:
<em>At equilibrium, the rate of the forward, and the reverse reactions are equal.</em>
Explanation:
In an equilibrium chemical reaction, the rate of forward reaction, is equal to the rate of reverse reaction. Note that the reactions does not cease at equilibrium, but rather, the reactants are converted to product, at the same rate at which the product is also being converted into the reactants in the reaction. When chemical equilibrium is reached, a careful calculation of the value of equilibrium constant is approximately equal to 1.
NB: If the value of equilibrium constant is far far greater than 1, then the reaction will favors more of the forward reaction, and if far far less than 1, the reaction will favor more of the reverse reaction.
