You have to use the equation PV=nRT.
P=pressure (in this case 1.89x10^3 kPa which equals 18.35677 atm)
1V=volume (in this case 685L)
n=moles (in this case the unknown)
R=gas constant (0.08206 (L atm)/(mol K))
T=temperature (in this case 621 K)
with the given information you can rewrite the ideal gas law equation as n=PV/RT.
n=(18.35677atm x 685L)/(0.08206atmL/molK x 621K)
n=246.8 moles
10, deca means ten a decagon is a ten side polygon, a decimetre is one tenth of a metre a decade is ten years so your answer is ten.
Answer:
13.73g
Explanation:
mass of reactants = mass of products.
Mass reactants = 5.00 g + 10.00 g = 15.00 g
Mass products = 1.27g + mass of ammonia and water vapor
Mass of ammonia and water vapor
15.00g – 1.27 g = 13.73 g
Answer:
![[base]=0.28M](https://tex.z-dn.net/?f=%5Bbase%5D%3D0.28M)
Explanation:
Hello,
In this case, by using the Henderson-Hasselbach equation one can compute the concentration of acetate, which acts as the base, as shown below:
![pH=pKa+log(\frac{[base]}{[acid]} )\\\\\frac{[base]}{[acid]}=10^{pH-pKa}\\\\\frac{[base]}{[acid]}=10^{4.9-4.76}\\\\\frac{[base]}{[acid]}=1.38\\\\](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29%5C%5C%5C%5C%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%3D10%5E%7BpH-pKa%7D%5C%5C%5C%5C%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%3D10%5E%7B4.9-4.76%7D%5C%5C%5C%5C%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%3D1.38%5C%5C%5C%5C)
![[base]=1.38[acid]=1.38*0.20M=0.28M](https://tex.z-dn.net/?f=%5Bbase%5D%3D1.38%5Bacid%5D%3D1.38%2A0.20M%3D0.28M)
Regards.
