Answer:
0.008 moles of gas are present
Explanation:
Given data:
Volume of gas = 1.75 L
Number of moles = ?
Temperature of gas = 58°C
Pressure of gas = 12.5 KPa
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Now we will convert the temperature.
58+273 = 331 K
Pressure = 12.5/101 = 0.12 atm
by putting values in formula:
0.12 atm× 1.75 L = n× 0.0821 atm.L/ mol.K ×331 K
0.21 atm.L = n× 27.17atm.L/ mol
n = 0.21 atm.L /27.17atm.L/ mol
n = 0.008 mol
Answer:
Al is oxidized while Ag is reduced.
Explanation:
The complete molecular equation is;
3Ag2S + 2Al --> 6Ag + Al2S3
Oxidation half equation;
2Al ------> 2Al^3+ + 6e
Reduction half equation;
6Ag^+ + 6e -------> 6Ag
Overall redox reaction equation;
2Al + 6Ag^+ ----->2Al^3+ + 6Ag
Hence; Al is oxidized while Ag is reduced.
Platinum isotope amu percent abundance
194
195 x 344
Missing 1x76
M=38.6 g
v=2 cm³
p=m/v
p=38.6/2=19.3 g/cm³
19.3 g/cm³
