First, we determine the number of moles of barium phosphate. This is done using:
Moles = mass / Mr
Moles = 1 / 602
Moles = 0.002
Now, we see from the formula of the compound that each mole of barium phosphate has 2 moles of phosphorus (P). Therefore, the moles of phosphorus are:
0.002 * 2 = 0.004
The number of particles in one mole of substance is 6.02 x 10²³. The number of phosphorus atoms will be:
0.004 * 6.02 x 10²³
2.41 x 10²¹ atoms of phosphorus are present
Answer:
Hydrogen, H_2
Explanation:
mass of each gas is 10.0 g
number of mole = mass/ molar mass
number of moles is directly proportional to volume at constant temp and pressure
this implies that the volume is inversely proportional to molar mass. And Among all the gases in periodic table the molar mass of Hydrogen is the least.
molar mass of H2=2 g/mol
Since, H2 has minimum molar mass then for the same mass of the gases Hydrogen will have maximum volume.
We can rephrase the statement with a little more specificity in order to understand the answer here.
The mass of the products can never be more than the The mass that is expected.
Answer: C. 25.6 kPa
Explanation:
The Gauge pressure is defined as the amount of pressure in a fluid that exceeds the amount of pressure in the atmosphere.
As such, the formula will be,
PG = PT – PA
Where,
PG is Gauge Pressure
PT is Absolute Pressure
PA is Atmospheric Pressure
Inputted in the formula,
PG = 125.4 - 99.8
PG = 25.6 kPa
The gauge pressure inside the container is 25.6kPa which is option C.
Answer : The cell emf for this cell is 0.118 V
Solution :
The half-cell reaction is:

In this case, the cathode and anode both are same. So,
is equal to zero.
Now we have to calculate the cell emf.
Using Nernest equation :
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}{diluted}]}{[Cl^{-}{concentrated}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BCl%5E%7B-%7D%7Bdiluted%7D%5D%7D%7B%5BCl%5E%7B-%7D%7Bconcentrated%7D%5D%7D)
where,
n = number of electrons in oxidation-reduction reaction = 1
= ?
= 0.0222 M
= 2.22 M
Now put all the given values in the above equation, we get:


Therefore, the cell emf for this cell is 0.118 V