What is the molar mass of Al^2(SO^4)3
1 answer:
You might be interested in
The ideal gas law may be written as
where
p = pressure
ρ =density
T = temperature
M = molar mass
R = 8.314 J/(mol-K)
For the given problem,
ρ = 0.09 g/L = 0.09 kg/m³
T = 26°C = 26+273 K = 299 K
M = 1.008 g/mol = 1.008 x 10⁻³ kg/mol
Therefore
Note that 1 atm = 101325 Pa
Therefore
p = 2.2195 x 10⁵ Pa
= 221.95 kPa
= (2.295 x 10⁵)/101325 atm
= 2.19 atm
Answer:
2.2195 x 10⁵ Pa (or 221.95 kPa or 2.19 atm)
where
p = pressure
ρ =density
T = temperature
M = molar mass
R = 8.314 J/(mol-K)
For the given problem,
ρ = 0.09 g/L = 0.09 kg/m³
T = 26°C = 26+273 K = 299 K
M = 1.008 g/mol = 1.008 x 10⁻³ kg/mol
Therefore
Note that 1 atm = 101325 Pa
Therefore
p = 2.2195 x 10⁵ Pa
= 221.95 kPa
= (2.295 x 10⁵)/101325 atm
= 2.19 atm
Answer:
2.2195 x 10⁵ Pa (or 221.95 kPa or 2.19 atm)
The oxidation is occurring on Calcium ions as it release one electron and reduction will be occurring on fluorine ion as it accepts one electron.
<u>Explanation:</u>
An element will undergo oxidation and form a positive ion on releasing one or more electrons from its valence shell. While reduction is occurred in a chemical reaction, then the element will be forming a negative ion with the acceptance of one or more electrons in its valence shell.
So in the given process of calcium fluoride, the one electron from the valence shell of calcium will be released making it as ions and this is termed as oxidation process. This one electron will be getting accepted by the fluorine ion and thus it will convert to
ions. This process of acceptance of electrons is termed as reduction.