Answer : The mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.
Explanation : Given,
Mass of oxygen in sulfur dioxide = 3.49 g
Mass of sulfur in sulfur dioxide = 3.50 g
Mass of oxygen in sulfur trioxide = 9.00 g
Mass of sulfur in sulfur trioxide = 6.00 g
Now we have to calculate the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide.
Mass of oxygen per gram of sulfur for sulfur dioxide = 
Mass of oxygen per gram of sulfur for sulfur dioxide = 
and,
Mass of oxygen per gram of sulfur for sulfur trioxide = 
Mass of oxygen per gram of sulfur for sulfur trioxide = 
Thus, the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.
<span> Ag(NH3)2Cl + 3HNO3 = AgNO3 +2NH4NO3 + HCl </span>
<span>or
Ag(NH3)2Cl + HNO3 = Ag(NH3)2NO3 + HCl this the complete balanced equation
now remove spectator ions to get net ionic equation
so
</span>
<span>
2H+ + 2NO3- + [Ag(NH3)2]+ Cl- -> AgCl + 2NH4+ + 2NO3- 2NO3- 2H+ [Ag(NH3)2]+ + Cl- -> AgCl + 2NH4+
</span>hope it helps
Answer: The balanced equation is
.
Explanation:
The given reaction equation is as follows.

Number of atoms present on reactant side are as follows.
- Li = 1
- H = 1
= 1
Number of atoms present on product side are as follows.
- Li = 1
- H = 2
= 1
To balance this equation, multiply Li by 2 and
by 2 on reactant side. Also, multiply
by 2 on product side.
Hence, the equation can be rewritten as follows.

Now, number of atoms present on reactant side are as follows.
- Li = 2
- H = 2
= 2
Number of atoms present on product side are as follows.
- Li = 2
- H = 2
= 2
As there are same number of atoms on both reactant and product side. Hence, the equation is now balanced.
Thus, we can conclude that the balanced equation is
.
976,563 is the answer I got to your question
Use the 1st order decay equation
<span><span>Cfinal</span>=<span>Cinitial</span><span>e<span>−kt</span></span></span>
by filling in and doing the math you will get:
<span> </span><span>Cfinal</span><span> = 1.11 grams Rn-222 after 23 days.
hope this helps
</span>