When the reaction equation is:
HgBr2(s) ↔ Hg2+(aq) + 2Br-(aq)
So Ksp expression = [Hg2+] [Br-]^2
assume the solubility S = X = 2.66 x 10^-7 M
and from the reaction equation :
we can see that [Hg2+] = X
and the [Br-] = 2 X
so by substitution in Ksp formula will can get the Ksp value:
∴ Ksp = X * (2X)^2
= 2.66 x 10^-7 * (2*2.66 x 10^-7)^2
= 7.53 x 10^-20
Answer:
The space occupy by the 180.0 g of gold will be 9.32 cm³
Explanation:
Given data:
Mass of gold = 180.0 g
Volume occupy by the gold = ?
Solution:
The given problem will be solved through the density formula.
Density is equal to the mass of substance divided by its volume.
Units:
SI unit of density is Kg/m3.
Other units are given below,
g/cm3, g/mL , kg/L
Formula:
D=m/v
D= density
m=mass
V=volume
The density of gold from literature is 19.320 g/cm³
Now we will put the values in formula.
19.320 g/cm³ = 180.0 g/ V
V = 180.0 g/ 19.320 g/cm³
V = 9.32 cm³
The space occupy by the 180.0 g of gold will be 9.32 cm³.
<u>Answer:</u> The volume of hydrogen gas required for the given amount of ethylene gas is 113 L
<u>Explanation:</u>
At STP:
1 mole of a gas occupies 22.4 L of volume
We are given:
Volume of ethylene = 113 L
For the given chemical equation:
By Stoichiometry of the reaction:
of ethylene reacts with of hydrogen gas
So, 113 L of ethylene gas will react with = 
of hydrogen gas
Hence, the volume of hydrogen gas required for the given amount of ethylene gas is 113 L
The given question is incomplete, the complete question is:
A reaction between liquid reactants takes place at 10.0 °c in a sealed, evacuated vessel with a measured volume of 35.0 L. Measurements show that the reaction produced 28. g of dinitrogen difluoride gas. Calculate the pressure of dinitrogen difluoride gas in the reaction vessel after the reaction. You may ignore the volume of the liquid reactants. Round your answer to 2 significant digits. pressure:atm
Answer:
The correct answer is 0.28 atm.
Explanation:
The temperature mentioned in the given reaction is 10 degree C, which after conversion becomes 283 Kelvin (273+10 = 283K).
The volume mentioned in the reaction is 35 Liters.
The reaction produced 28 grams of dinitrogen difluoride gas (N2F2). The n or the no of moles of the gas can be determined with the help of the formula:
moles of N2F2 = mass/molar mass
= 28/66 (molar mass of N2F2 is 66 g/mol)
= 0.424
The pressure of the gas can be determined by using the equation of the ideal gas law, that is, PV = nRT
P * 35 = 0.424 * 0.0821 * 283
P = 0.28 atm
