Answer:
A or C because they both kinda are the same
Answer: 1.98 g
Explanation:
To calculate the moles :
The balanced given equation is:
According to stoichiometry :
4 moles of
will produce = 2 moles of
Thus 0.22 moles of
will produce=
of
Mass of
Thus 1.98 g of water is produced along with 5.0 L of
at STP
Explanation:
The given data is as follows.
Volume of lake =
= 
Concentration of lake = 5.6 mg/l
Total amount of pollutant present in lake = 
=
mg
=
kg
Flow rate of river is 50 
Volume of water in 1 day = 
=
liter
Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are
or 
Flow rate of sewage = 
Volume of sewage water in 1 day =
liter
Concentration of sewage = 300 mg/L
Total amount of pollutants =
or 
Therefore, total concentration of lake after 1 day = 
= 6.8078 mg/l
= 0.2 per day
= 6.8078
Hence,
= 
=
= 1.234 mg/l
Hence, the remaining concentration = (6.8078 - 1.234) mg/l
= 5.6 mg/l
Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.
Answer: 6.Explanation:1) Aluminum

So each atom of aluminum lost 3 electrons to pass from 0 oxidation state to 3+ oxidation state.
2) Manganesium

So, each ion of Mn(2+) gained 2 electrons pass from 2+ oxidation state to 0.
3) Balance
Multiply aluminum half-reaction (oxidation) by 2 and multiply manganesium half-raction (reduction) by 3:

4) Net equation
Add the two half-equations:

As you see the left side has 2 Al, 3Mn, and 3*2 positive charges.
The right side has 2 Al, 3 Mn, and 2*3 positive charges.
So, the equation is balanced.
5) Count the number of electrons involved.
As you see 2 atoms of aluminum lost 6 electrons (3 each).
That is the answer to the question. 6 electrons will be lost.
Im pretty sure the answer is 4 but not 100 percent