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:
3 moles
Explanation:
SrCO3
Mass = 442.8g
Molar mass = (87.6 + 12 * [16*3]) = 147.6g/mol
Number of moles = mass / molar mass
Number of moles = 442.8 / 147.6
Number of moles = 3
Answer:0.8742j/g°C
Explanation: SOLUTION
GIVEN
length of bar=1.25m
mass 382g
temperature= 20°C to 288°C
Q=89300J
Specific Heat Capacity will be calculated using
Q=mC∆T
where
C = specific heat capacity
Q = heat
m = mass
Δ T = change in temperature
C=Q/ m∆T
=89300/382X(288-20.6)
=0.8742j/g°C
Answer:
Explanation:
We are given that 25 mL of 0.10 M 
is titrated with 0.10 M NaOH(aq).
We have to find the pH of solution
Volume of 
Volume of NaoH=0.01 L
Volume of solution =25 +10=35 mL=
Because 1 L=1000 mL
Molarity of NaOH=Concentration OH-=0.10M
Concentration of H+= Molarity of 
=0.10 M
Number of moles of H+=Molarity multiply by volume of given acid
Number of moles of H+=
=0.0025 moles
Number of moles of 
=0.001mole
Number of moles of H+ remaining after adding 10 mL base = 0.0025-0.001=0.0015 moles
Concentration of H+=
pH=-log [H+]=-log [4.28
]=-log4.28+2 log 10=-0.631+2
