Answer:
2HgS + 3O2 → 2HgO + 2SO2
The coefficients are: 2, 3, 2, 2
Explanation:
HgS + O2 → HgO + SO2
The equation can be balance as follow:
Put 3 in front of O2 as shown below:
HgS + 3O2 → HgO + SO2
Now we can see that there are 6 atoms of O on the left side of the equation and a total of 3 atoms on the right side. It can be balance by putting 2 in front of HgO and SO2 as shown below:
HgS + 3O2 → 2HgO + 2SO2
Now we have 2 atoms of both Hg and S on the right side and 1atom each on the left. It can be balance by putting 2 in front of HgS as shown below:
2HgS + 3O2 → 2HgO + 2SO2
Now the equation is balanced.
The coefficients are: 2, 3, 2, 2
The law of conservation of mass(matter) states that matter(mass) can neither be created nor destroyed during a chemical reaction but changes from one form to another. An unbalanced equation suggests that matter has been created or destroyed. While a balanced equation proofs that matter can never be created but changes to different form. This is the more reason we have count the atoms of an element on both side of the equation to see if they are balanced irrespective of the new form they assume in the product
Answer:
Total 5 significant digits.
Explanation:
Significant digits are the numbers that give a meaningful contribution. For example, digit 013 has the 2 significant digits and zero is not a significant digit because digits 1 and 3 give meaningful contribution but digit zero does not value meaningful contribution. Similarly, the 89015 has a total of 5 significant digits and these digits are the 8, 9, 0, 1, and 5.
Answer: 116 g of copper
Explanation:
where Q= quantity of electricity in coloumbs
I = current in amperes = 24.5A
t= time in seconds = 4.00 hr = 
(1hr=3600s)
of electricity deposits 63.5 g of copper.
352800 C of electricity deposits = 
of copper.
Thus 116 g of Cu(s) is electroplated by running 24.5A of current
Thus remaining in solution = (0.1-0.003)=0.097moles
I honestly have no clue on this one
You have 0.50 mol of NH3 and 0.20 mol of NH4+ to start (NH4Cl dissolves completely), given the molarity and 1.0 L solution.
30.0 mL of 1.0 M HCl is 0.0300 mol of HCl. This will react with the NH3 to produced 0.030 mol of NH4+.
You now have 0.47 mol NH3 and 0.23 mol NH4+. Now use the Henderson-Hasselbach equation to calculate your pH. The equation says to use concentration of acid and base, but you can just use the moles of them because it doesn’t make a difference.
pH = pKa + log(base/acid)
pKa = 14 - pKb = 14 - 4.75 = 9.25
pH = 9.25 + log(0.47/0.23) = 9.56
