Answer: Option (B) is the correct answer.
Explanation:
A covalent compound is a compound formed by sharing of electrons. And, in a covalent network solid atoms are bonded by covalent bonds in a continuous network that is extending throughout the material or solid.
This continuous arrangement of atoms are like a lattice.
For example, diamond is a covalent network solid in which carbon atoms are arranged in a continuous lattice like structure.
Hence, we can conclude that the statement all the atoms are covalently bonded to other atoms to form a lattice-like structure, best describes the structure of covalent network solids.
Answer:
Pyrophoricity is a property of metals and oxides of lower oxidation states, including radioactive ones, in which they spontaneously ignite during or after stabilization.
Answer:
Equilibrium concentration of
is 12.5 M
Explanation:
Given reaction: 
Here, ![K_{c}=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}][H_{2}O]}](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5BC_%7B2%7DH_%7B5%7DOH%5D%7D%7B%5BC_%7B2%7DH_%7B4%7D%5D%5BH_%7B2%7DO%5D%7D)
where
represents equilibrium constant in terms of concentration and species inside third bracket represent equilibrium concentrations
Here,
,
and 
So, ![[H_{2}O]=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}]\times K_{c}}=\frac{1.69}{0.015\times 9.0}=12.5M](https://tex.z-dn.net/?f=%5BH_%7B2%7DO%5D%3D%5Cfrac%7B%5BC_%7B2%7DH_%7B5%7DOH%5D%7D%7B%5BC_%7B2%7DH_%7B4%7D%5D%5Ctimes%20K_%7Bc%7D%7D%3D%5Cfrac%7B1.69%7D%7B0.015%5Ctimes%209.0%7D%3D12.5M)
Hence equilibrium concentration of
is 12.5 M
Answer:
378.25°C
Explanation:
Given data:
Number of moles of gas = 0.750 mol
Volume of gas = 35.9 L
Pressure of gas = 114 KPa (114/101 = 1.125 atm)
Temperature of gas = ?
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 put the values.
T = PV/nR
T = 1.125 atm × 35.9 L /0.750 mol × 0.0821 atm.L/ mol.K
T = 40.3875/0.062/K
T = 651.4 K
Kelvin to °C:
651.4 K - 273.15 = 378.25°C
Mole percent of O2 = 10% = 0.1 moles
Mole percent of N2 = 10% = 0.1 moles
Mole percent of He = 80% = 0.8 moles
Molar Mass of O2 = (2 x 16) x 0.1 = 3.2
Molar Mass of N2 = (2 x 14) x 0.1 = 2.8
Molar Mass of He = 4 x 0.8 = 3.2
1. Molar Mass of the mixture = 3.2 + 2.8 + 3.2 = 9.2 grams
2. Since at constant volume density is proportional to mass, so the ratio of
mass will be the ratio of density.
Ratio = Molar Mass of the mixture / Molar Mass of O2 = 9.2 / 32 = 0.2875