Answer:
Vol of 4 moles CO₂(g) at STP = 89.6 Liters
Explanation:
STP
P = 1 Atm
V =
T = 0°C = 273 K
n = 4 moles
R = 0.08206 L·Atm/mol·K
Using Ideal Gas Law PV = nRT => V = nRT/P
V = (4 moles)(0.08206 L·Atm/mol·K)(273 K)/(1 Atm) = 89.6 Liters
Answer:

Explanation:
The molar mass of uranium-235 is 235 g/mol. So one mole of uranium-235 has a mass of 235 g. Put differently 6.022×10^23 atoms of uranium-235 have a mass of 235 g. Knowing that, how can we use that to find the mass of one atom?
mass of one atom = 
Answer:
4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.
Explanation:
By ideal gas equation:

Number of moles (n)
can be written as: 
where, m = given mass
M = molar mass

where,
which is known as density of the gas
The relation becomes:
.....(1)
We are given:
M = molar mass of chloroform= 119.5 g/mol
R = Gas constant = 
T = temperature of the gas = 
P = pressure of the gas = 1.00 atm
Putting values in equation 1, we get:

4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.
Answer:
The correct option is;
4 percent ionic, 96 percent covalent, 222 pm
Explanation:
The parameters given are;
Phosphorus:
Atomic radius = 109 pm
Covalent radius = 106 pm
Ionic radius = 212 pm
Electronegativity of phosphorus = 2.19
Selenium:
Atomic radius = 122 pm
Covalent radius = 116 pm
Ionic radius = 198 pm
Electronegativity of selenium= 2.55
The percentage ionic character of the chemical bond between phosphorus and selenium is given by the relation;
Using Pauling's alternative electronegativity difference method, we have;
![\% \, Ionic \ Character = \left [18\times (\bigtriangleup E.N.)^{1.4} \right ] \%](https://tex.z-dn.net/?f=%5C%25%20%5C%2C%20Ionic%20%5C%20Character%20%3D%20%5Cleft%20%5B18%5Ctimes%20%28%5Cbigtriangleup%20E.N.%29%5E%7B1.4%7D%20%20%5Cright%20%5D%20%5C%25)
Where:
Δ E.N. = Change in electronegativity = 2.55 - 2.19 = 0.36
Therefore;
![\% \, Ionic \ Character = \left [18\times (0.36)^{1.4} \right ] \% = 4.3 \%](https://tex.z-dn.net/?f=%5C%25%20%5C%2C%20Ionic%20%5C%20Character%20%3D%20%5Cleft%20%5B18%5Ctimes%20%280.36%29%5E%7B1.4%7D%20%20%5Cright%20%5D%20%5C%25%20%3D%204.3%20%5C%25)
Hence the percentage ionic character = 4.3% ≈ 4%
the percentage covalent character = (100 - 4.3)% = 95.7% ≈ 96%
The bond length for the covalent bond is found adding the covalent radii of both atoms as follows;
The bond length for the covalent bond = 106 pm + 116 pm = 222 pm.
The correct option is therefore, 4 percent ionic, 96 percent covalent, 222 pm.