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
Answer:
27 liters of hydrogen gas will be formed
Explanation:
Step 1: Data given
Number of moles C = 1.03 moles
Pressure H2 = 1.0 atm
Temperature = 319 K
Step 2: The balanced equation
C +H20 → CO + H2
Step 3: Calculate moles H2
For 1 mol C we need 1 mol H2O to produce 1 mol CO an 1 mol H2
For 1.03 moles C we'll have 1.03 moles H2
Step 4: Calculate volume H2
p*V = n*R*T
⇒with p = the pressure of the H2 gas = 1.0 atm
⇒with V = the volume of H2 gas = TO BE DETERMINED
⇒with n = the number of moles H2 gas = 1.03 moles
⇒with R = the gas constant = 0.08206 L*Atm/mol*K
⇒with T = the temperature = 319 K
V = (n*R*T)/p
V = (1.03 * 0.08206 *319) / 1
V = 27 L
27 liters of hydrogen gas will be formed
Its c ok im sorry if i got the last answer wrong
Answer : The final temperature of the mixture is 
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
where,
= specific heat of iron = 
= specific heat of water = 
= mass of iron = 39.9 g
= mass of water = 
= final temperature of mixture = ?
= initial temperature of iron = 
= initial temperature of water = 
Now put all the given values in the above formula, we get
Therefore, the final temperature of the mixture is
