2Cu(NO3)2 ----------> 2CuO (s) +4NO2 (g) + O2(g)
9.378g=0.05moles
no of moles = weight / MW = 9.378/187.56 = 0.05moles
as per the above reaaction 2moles of Cu(NO3)2 can produce 4moles of N2
0.05moles Cu(NO3)2 can produce (0.05*4)/2 = 0.1moles of N2
and 2moles of Cu(NO3)2 can produce 1moles of O2
0.05moles Cu(NO3)2 can produce (0.05*1)/2 = 0.025moles of O2
Total moles of gas i.e., N2 and O2 =0.1+0.025 = 0.125moles
From PV = nRT
V = nRT/ P = 0.125*0.0821*273 = 2.80166Lit option is correct
Answer: sand, silt, and clay.
Answer:
Ok:
Explanation:
So, you can use the Henderson-Hasselbalch equation for this:
pH = pKa + log(
) where A- is the conjugate base of the acid. In other words, A- is the deprotonated form and HA is the protonated.
We can solve that
1 = log(
) and so 10 = or 10HA = A-. For every 1 protonated form of adenosine (HA), there are 10 A-. So, the percent in the protonated form will be 1(1+10) or 1/11 which is close to 9 percent.
D increase in temperature and increase in pressure.
