Answer:
U= 238g/mol
U2O5= 556g/mol
Explanation:
Since U= 238
O=16
U3O5= 2(238)+3(16)=556g/mol
Answer:
Therefore the equilibrium number of vacancies per unit cubic meter =2.34×10²⁴ vacancies/ mole
Explanation:
The equilibrium number of of vacancies is denoted by 
.
It is depends on
- total no. of atomic number(N)
- energy required for vacancy
- Boltzmann's constant (k)= 8.62×10⁻⁵ev K⁻¹
- temperature (T).
To find equilibrium number of of vacancies we have find N.
Here ρ= 8.45 g/cm³ =8.45 ×10⁶m³
= Avogadro Number = 6.023×10²³
= 63.5 g/mole
g/mole
Here 
=0.9 ev/atom , T= 1000k
Therefore the equilibrium number of vacancies per unit cubic meter,
=2.34×10²⁴ vacancies/ mole
<span>The pressure inside a coke bottle is really high. This helps keep the soda carbonated. That is, the additional pressure at the surface of the liquid inside the bottle forces the bubbles to stay dissolved within the soda. </span><span>When the coke is opened, there is suddenly a great pressure differential. The initial loud hiss that is heard is this pressure differential equalizing itself. All of the additional pressure found within the bottle pushes gas out of the bottle until the pressure inside the bottle is the same as the pressure outside the bottle. </span><span>However, once this occurs, the pressure inside the bottle is much lower and the gas bubbles that had previously been dissolved into the soda have nothing holding them in the liquid anymore so they start rising out of the liquid. As they reach the surface, they pop and force small explosions of soda. These explosions are the source of the popping and hissing that continues while the soda is opened to the outside air. Of course, after a while, the soda will become "flat" when the only gas left dissolved in the liquid will be the gas that is held back by the relatively weak atmospheric pressure.</span>
Answer:
Option C. 4.03 g
Explanation:
Firstly we analyse data.
12 % by mass, is a sort of concentration. It indicates that in 100 g of SOLUTION, we have 12 g of SOLUTE.
Density is the data that indicates grams of solution in volume of solution.
We need to determine, the volume of solution for the concentration
Density = mass / volume
1.05 g/mL = 100 g / volume
Volume = 100 g / 1.05 g/mL → 95.24 mL
Therefore our 12 g of solute are contained in 95.24 mL
Let's finish this by a rule of three.
95.24 mL contain 12 g of sucrose
Our sample of 32 mL may contain ( 32 . 12) / 95.24 = 4.03 g
