The two half-reactions are...
Ag→Ag+
and...
NO3→NO
Let's start by balancing the first half-reaction...
Ag→Ag+
The amounts are already balanced; 1:1. The oxygens are balanced. So all that's left is to balance the charge...
Ag→Ag++e−
Now let's do the other equation... Amounts of nitrogen are balanced, so we first need to balance the oxygens...
NO3→NO
4H++NO3→NO+2H2O
Next, we need to balance charge...
4e−+4H++NO3→NO+2H2O
Now let's go ahead and rewrite each half-reaction after being balanced by themselves...
Ag→Ag++e−
4e−+4H++NO3→NO+2H2O
Now we need to multiply by some factor to get the electrons to cancel out. In this case, that factor is 4, which needs to be applied to the top half-reaction...
4(Ag→Ag++e−)=4Ag→4Ag++4e−
Then we combine this half-reaction with the second one above to get...
4Ag+4H++NO3→4Ag++NO+2H2O
B 2
is which shell are the valence electrons of the elements in period 2 found
Answer : The mass defect required to release energy is 6111.111 kg
Explanation :
To calculate the mass defect for given energy released, we use Einstein's equation:
E = Energy released = 
= mass change = ?
c = speed of light = 
Now put all the given values in above equation, we get:
Therefore, the mass defect required to release energy is 6111.111 kg
Answer:
The change in temperature that occurs when 8000 J of heat is used by a mass 75 g of water is 25.4 °C
Explanation:
H = mc ΔT
m = 75 g
c = 4. 200 J/ g °C
H = 8000 J
ΔT =?
Rearranging the formula, making ΔT the subject of formula;
ΔT = H / m c
ΔT = 8000 / 75 * 4.200
ΔT = 8000 / 315
ΔT = 25.4 °C
The answer is B. Below freezing
