A 0.200 M of K2SO4 solution is produced by diluting 20.0 mL of 5.00 M K2SO4 solution to 500.0 mL.
<u>Explanation</u>:
- When dealing with dilution we will use the following equation:
M1 V1 = M2 V2
where,
M1 = initial concentration
V1 = initial volume
M2 = final concentration
V2 = final volume
- By diluting 20.0 mL of 5.00 M K2SO4 solution to 500.0 mL, we get
M1 V1 = M2 V2
20.0 mL 
5.00 M = M2 500.0 mL
M2 = (20.0 mL 5.00 M) / 500.0 mL
M2 = 0.200 M.
Hence A 0.200 M of K2SO4 solution is produced by diluting 20.0 mL of 5.00 M K2SO4 solution to 500.0 mL.
Answer:
See explanation
Explanation:
Now , we have the equation of the reaction as;
2H2S(g) + 302(g)------->2SO2(g) + 2H2O(g)
This equation shows that SO2 gas is produced in the process. Let us recall that this same SO2 gas is the anhydride of H2SO4. This means that it can dissolve in water to form H2SO4
So, when SO2 dissolve in rain droplets, then H2SO4 is formed thereby lowering the pH of rain water. This is acid rain.
The mean is simply the arithmetic average of all your raw data. This can be solved methodically by summing up all of the raw data points that you have. Take note how many raw data points you used, because this will be used to divide the sum. You will obtain the mean.
The net cell reaction for the iron-silver voltaic cell would be:
<span>3Ag+ + Fe --------> 3Ag + Fe3+
</span>
wherein it comes from the reaction of the cathode and the anode:
<span>Ag+ + e- -------> Ag, for the cathode and;
</span>
<span>Fe -------> Fe3+ + 3e-, for the anode
</span>
When both equations would be multiplied, the it would now yield the balanced net reaction as stated above.
The solution for this problem would be:
We are looking for the grams of magnesium that would have
been used in the reaction if one gram of silver were created. The computation
would be:
1 g Ag (1 mol Mg) (24.31 g/mol) / (2mol Ag)(107.87g/mol) =
0.1127 grams of Magnesium
