Answer:
What mass (g) of barium iodide is contained in 188 mL of a barium iodide solution that has an iodide ion concentration of 0.532 M?
A) 19.6
B) 39.1
C) 19,600
D) 39,100
E) 276
The correct answer to the question is
B) 39.1 grams
Explanation:
To solve the question
The molarity ratio is given by
188 ml of 0.532 M solution of iodide.
Therefore we have number of moles = 0.188 × 0.532 M = 0.100016 Moles
To find the mass, we note that the Number of moles =
from which we have
Mass = Number of moles × molar mass
Where the molar mass of Barium Iodide = 391.136 g/mol
= 0.100016 moles ×391.136 g/mol = 39.12 g
Answer : The cell emf for this cell is 0.118 V
Solution :
The half-cell reaction is:

In this case, the cathode and anode both are same. So,
is equal to zero.
Now we have to calculate the cell emf.
Using Nernest equation :
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}{diluted}]}{[Cl^{-}{concentrated}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BCl%5E%7B-%7D%7Bdiluted%7D%5D%7D%7B%5BCl%5E%7B-%7D%7Bconcentrated%7D%5D%7D)
where,
n = number of electrons in oxidation-reduction reaction = 1
= ?
= 0.0222 M
= 2.22 M
Now put all the given values in the above equation, we get:


Therefore, the cell emf for this cell is 0.118 V
Answer:
The average atomic mass of bromine is 79.9 amu.
Explanation:
Given data:
Percentage of Br⁷⁹ = 55%
Percentage of Br⁸¹ = 45%
Average atomic mass of bromine = ?
Formula:
Average atomic mass = [mass of isotope× its abundance] + [mass of isotope× its abundance] +...[ ] / 100
Now we will put the values in formula.
Average atomic mass = [55 × 79] + [81 ×45] / 100
Average atomic mass = 4345 + 3645 / 100
Average atomic mass = 7990 / 100
Average atomic mass = 79.9 amu
The average atomic mass of bromine is 79.9 amu.
Answer:
The coefficient of O2 is 11
Explanation:
Step 1:
The equation for the reaction:
FeS2 + O2 → SO2 + Fe2O3
Step 2:
Balancing the equation. The equation can be balance as follow:
FeS2 + O2 → SO2 + Fe2O3
There are 2 atoms of Fe on the right side and 1 atom on the left. It can be balance by putting 2 in front of FeS2 as shown below:
2FeS2 + O2 → SO2 + Fe2O3
There are 4 atoms of S on the left side and 1 atom on the right side. It can be balance by putting 4 in front of SO2 as shown below:
2FeS2 + O2 → 4SO2 + Fe2O3
Now, there are a total of 11 atoms of O on the right side and 2 atoms on the left side. It can be balance by putting 11/2 in front of O2 as shown below:
2FeS2 + 11/2O2 → 4SO2 + Fe2O3
Multiply through by 2 to clear the fraction as shown below:
4FeS2 + 11O2 → 8SO2 + 2Fe2O3
Now the equation is balanced.
The coefficient of O2 is 11
Explanation:
Dichloromethane is flammable - FALSE
Methanol is flammable. - TRUE
Concentrated sulfuric acid is corrosive. - TRUE
10% sodium carbonate solution must be used in the fume hood. - FALSE
Benzoyl chloride is a lachrymator. - TRUE