B. 272
I think because it's 3 sig figs
<span>ZnS + O --> ZnO + SO
Okay so first you have to count up the number of elements on each side of the equation. Your objective is to have the same number of each element on both sides.
Left Side:
Zn - 1
S - 1
O - 1
Right Side:
Zn - 1
S - 1
O - 2
Since there are two oxygens on the right side, you have to add a coefficient of 2 to the oxygen on the left side. The coefficient tells us that that element or molecule is being multiplied by the value of coefficient. Since we're adding a coefficient of 2 to the oxygen on the left side, there are now 2 oxygens on that side. Because that is the same amount of oxygen as on the right, the equation is now balanced.
Your final equation should look like this: </span>ZnS + 2O --> ZnO + SO
In conclusion, the answer is 2.
Mghcl14 + H2 = mghclh 16 work needed to solve this problem
Reduction reactions are those reactions that reduce the oxidation number of a substance. Hence, the product side of the reaction must contain excess electrons. The opposite is true for oxidation reactions. When you want to determine the potential difference expressed in volts between the cathode and anode, the equation would be: E,reduction - E,oxidation.
To cancel out the electrons, the e- in the reactions must be in opposite sides. To do this, you reverse the equation with the negative E0, then replacing it with the opposite sign.
Pb(s) --> Pb2+ +2e- E0 = +0.13 V
Ag+ + e- ---> Ag E0 = +0.80 V
Adding up the E0's would yield an overall electric cell potential of +0.93 V.
Answer:
3 significant zeroes
Explanation:
To count the number of significant figures, you must pass the zeroes until you reach a non-zero value. Once you reach it, count anything after that as significant values, including the non-zero value itself.
The number has 4 significant figures with 3 significant zeroes.
Hope this helps!!!
