When 1.95 g of co(no3)2 is dissolved in 0.350 l of 0.220 m koh, what are [ co2+], [ co(oh)42−], and [ oh−] if kf of co(oh)42− =
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<span>Uranium-236 is intermediate nuclei. created by fusion reactions an unstable isotope of uranium created from four hydrogen atoms used in the H-bomb.
Following is the reaction involved in above process:
</span>
+ 
→ 
→ 
+ 
+ 3 <span> + 177 MeV
</span>
Here,
= Fission material,
= projectile,
= intermediate nuclei,
and = Fission product
Following is the reaction involved in above process:
</span>
</span>
Here,
Answer:
A. fluorine, 1.79 moles
Explanation:
Given parameters:
Mass of carbon = 87.7g
Mass of fluorine gas = 136g
Unknown:
The limiting reactant and the maximum amount of moles of carbon tetrafluoride that can be produced = ?
Solution:
Equation of the reaction:
C + 2F₂ → CF₄
let us find the number of the moles the given species;
Number of moles =
C; molar mass = 12;
Number of moles = = 7.31moles
F; molar mass = 2(19) = 38g/mol
Number of moles = = 3.58moles
So;
From the give reaction:
1 mole of C requires 2 moles of F₂
7.31 moles of C will then require 2 x 7.31 moles of F₂ = 14.62moles
But we have 3.58 moles of the F₂;
Therefore, the reactant in short supply is F₂ and it is the limiting reactant;
So;
2 moles of F₂ will produce mole of CF₄
3.58 moles of F₂ will then produce = 1.79moles of CF₄
Answer : The number of moles of solute is, 0.0788 moles.
Explanation : Given,
Molarity = 0.225 M
Volume of solution = 0.350 L
Formula used:
Now put all the given values in this formula, we get:
Therefore, the number of moles of solute is, 0.0788 moles.