Answer:
The rate is a mathematical relationship obtained by comparing reaction rate with reactant concentrations.
Answer: 8
Explanation: Radon has 8 valence electrons. Radon is considered stable with a complete octet of electrons, filling the s and p orbitals.
Answer:

Explanation:
Hello!
In this case, when we want to balance chemical reactions such as in this case, the idea is to equal to number of atoms of each element at each side of the equation according to the lay of conservation of mass, just as shown below:

Because we have four phosphorous and ten oxygen atoms at each side.
Best regards!
The balanced chemical
reaction will be:
2H2O = 2H2 + O2
<span>We are given the amount of water used in the decomposition reaction. This will be our
starting point.</span>
<span>17.0 g H2O</span> (1 mol H2O/ 18.02 g H2O) (1 mol O2/2
mol <span>H2O</span>) ( 32.00 g O2/1mol O2) = 15.09 g O2
Percent yield = actual yield / theoretical yield x 100
<span>Percent yield =10.2 g / 15.09 g
x 100</span>
Percent yield = 67.58%
Answer : The concentration of
is, 
Explanation :
When we assume this reaction is driven to completion because of the large excess of one ion then we are assuming limiting reagent is
and
is excess reagent.
First we have to calculate the moles of KSCN.


Moles of KSCN = Moles of
= Moles of
= 
Now we have to calculate the concentration of ![[Fe(SCN)]^{2+}](https://tex.z-dn.net/?f=%5BFe%28SCN%29%5D%5E%7B2%2B%7D)
![\text{Concentration of }[Fe(SCN)]^{2+}=\frac{\text{Moles of }[Fe(SCN)]^{2+}}{\text{Volume of solution}}](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7D%5BFe%28SCN%29%5D%5E%7B2%2B%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7D%5BFe%28SCN%29%5D%5E%7B2%2B%7D%7D%7B%5Ctext%7BVolume%20of%20solution%7D%7D)
Total volume of solution = (6.00 + 5.00 + 14.00) = 25.00 mL = 0.025 L
![\text{Concentration of }[Fe(SCN)]^{2+}=\frac{1.08\times 10^{-5}mol}{0.025L}=4.32\times 10^{-4}M](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7D%5BFe%28SCN%29%5D%5E%7B2%2B%7D%3D%5Cfrac%7B1.08%5Ctimes%2010%5E%7B-5%7Dmol%7D%7B0.025L%7D%3D4.32%5Ctimes%2010%5E%7B-4%7DM)
Thus, the concentration of
is, 