Answer:
adding silver ions , adding chloride ions, removing chloride ions and removing silver chloride.
Explanation:
Barium sulfate is an insoluble ionic compound since it is insoluble in water.
We can differentiate barium ions and sulfate ions away from the surface if we add a small amount of solid barium sulfate is mixed with water and shaken.
The action would shift this reaction away from solid barium sulfate and toward the dissolved ions are adding silver ions , adding chloride ions, removing chloride ions and removing silver chloride.
An ionic and covalent quality has in every molecular bond.
In barium sulfate barium has lowest electronegativities.
Oxygen has highest electronegativities.
so the bond between both barium and sulfate is ionic in character.
second compound
Let molar mass of x is = X
Let molar mass of y is = Y
Moles of x in second compound = Mass / molar mass = 7 / X
Moles of y in second compound = Mass / molar mass = 4.5 / Y
For second compound
7 / X : 4.5/ Y = 1:1
Therefore
X / Y = 7/4.5
Y / X = 4.5/ 7
The mass of x in first compound = 14g
moles of x in first compound = 14/X
Mass of y in first compound = 3
moles of y in first compound = 3 / Y
14 / X : 3/ Y = 14Y / 3X = 14 X 4.5 / 3 X 7 = 3 :1
Thus molar ratio in first compound = moles of x / Moles of y = 3:2
Formula = x3y
Answer: The temperature of 0.6 moles of fluorine that occupy 15 L at 2,300 mmHg is 920 K
Explanation:
According to ideal gas equation:
P = pressure of gas = 2300 mm Hg = 3.02 atm (760mmHg=1atm)
V = Volume of gas = 15 L
n = number of moles = 0.6
R = gas constant =
T =temperature = ?
Thus the temperature of 0.6 moles of fluorine that occupy 15 L at 2,300 mmHg is 920 K
Answer : The correct option is, (b) +0.799 V
Solution :
The values of standard reduction electrode potential of the cell are:
![E^0_{[H^{+}/H_2]}=+0.00V](https://tex.z-dn.net/?f=E%5E0_%7B%5BH%5E%7B%2B%7D%2FH_2%5D%7D%3D%2B0.00V)
![E^0_{[Ag^{+}/Ag]}=+0.799V](https://tex.z-dn.net/?f=E%5E0_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D%3D%2B0.799V)
From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.
The half reaction will be:
Reaction at anode (oxidation) : 
Reaction at cathode (reduction) : 
The balanced cell reaction will be,
Now we have to calculate the standard electrode potential of the cell.
![E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}](https://tex.z-dn.net/?f=E%5Eo_%7Bcell%7D%3DE%5Eo_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D-E%5Eo_%7B%5BH%5E%7B%2B%7D%2FH_2%5D%7D)
Therefore, the standard cell potential will be +0.799 V
A division or split
in biology a cell division
