<span>1.40 x 10^5 kilograms of calcium oxide
The reaction looks like
SO2 + CaO => CaSO3
First, determine the mass of sulfur in the coal
5.00 x 10^6 * 1.60 x 10^-2 = 8.00 x 10^4
Now lookup the atomic weights of Sulfur, Calcium, and Oxygen.
Sulfur = 32.065
Calcium = 40.078
Oxygen = 15.999
Calculate the molar mass of CaO
CaO = 40.078 + 15.999 = 56.077
Since 1 atom of sulfur makes 1 atom of sulfur dioxide, we don't need the molar mass of sulfur dioxide. We merely need the number of moles of sulfur we're burning. divide the mass of sulfur by the atomic weight.
8.00 x 10^4 / 32.065 = 2.49 x 10^3 moles
Since 1 molecule of sulfur dioxide is reacted with 1 molecule of calcium oxide, just multiply the number of moles needed by the molar mass
2.49 x 10^3 * 56.077 = 1.40 x 10^5
So you need to use 1.40 x 10^5 kilograms of calcium oxide per day to treat the sulfur dioxide generated by burning 5.00 x 10^6 kilograms of coal with 1.60% sulfur.</span>
Hey there! :D
To some extent, all moving parts produce friction. It can be very small or minimal, but all parts produce some form of friction.
This is true.
I hope this helps!
~kaikers
First identify which is being oxidized and reduced. In this case, the Mg is being oxidized and the Hg is being reduced.
Mg --> Mg+2
<span>Hg+2 --> Hg+1
</span>
Then you have to balance each half reaction first with electrons before adding them together in one equation

⇒

and

⇒
and then combine them together to form

⇒

It isn't necessary to keep the electrons but its essential to know how many there are in order to know how many are in the equation in order to calculate the reaction energy. Note: A<span>dd H+ and H2O to balance the H's and O's in acidic solution if needed.</span>
Buffer solution resist the change in pH upon addition of small amount of strong acid or strong base.
Buffer consists of weak acid as HF / and its conjugate base NaF
When strong acid as HCl is added to buffer, it respond with its conjugate base to convert the strong acid to weak acid like this:
HCl (S.A) + NaF → NaCl + HF (W.A)
moles of HF we already have = M * V(in liters)
= 0.0955 M * 0.033 L = 3.15 x 10⁻³ mole
moles of HCl added = 8.00 x 10⁻⁵ mole
one mole HCl reacts with 1 mole NaF to give 1 mole HF
so the amount added to HF = 8.00 x 10⁻⁵
Total moles of HF present = (3.15 x 10⁻³) + (8.00 x 10⁻⁵) = 3.23 x 10⁻³ mole