Answer:
I think its B im not sure
but i hope this helps
The volume in liters occupied by 22.6 g of I₂ gas at STP is 1.99 L (answer A)
<u><em>calculation</em></u>
Step: find the moles of I₂
moles= mass÷ molar mass
from periodic table the molar mass of I₂ is 253.8 g/mol
moles = 22.6 g÷253.8 g/mol =0.089 moles
Step 2:find the volume of I₂ at STP
At STP 1 moles =22.4 L
0.089 moles= ? L
<em>by cross multiplication</em>
={ (0.089 moles x 22.4 L) /1 mole} = 1.99 L
The net amount of energy produced can be obtained from a table of enthalpy change of formation, available online.
The enthalpy change of formation indicate how much energy the 1 mole of the product (H2O) has relative to the elemental reactants (H2 and O2). In other words, the "lost" energy equals the heat/energy released.
For water (H2O), this value is -285.8 if the final product is a liquid under standard conditions, and -241.82 if the product is in gas form which contains some energy that could be further released. This means that if the final product (H2O) is in liquid form, energy released is 285.8 kJ/mol.
Since water is in liquid form under standard conditions, the first value (285.8 kJ/mol) is generally appropriate.
We will use Arrehenius equation
lnK = lnA -( Ea / RT)
R = gas constant = 8.314 J / mol K
T = temperature = 25 C = 298 K
A = frequency factor
ln A = ln (1.5×10 ^11) = 25.73
Ea = activation energy = 56.9 kj/mol = 56900 J / mol
lnK = 25.73 - (56900 / 8.314 X 298) = 2.76
Taking antilog
K = 15.8
Answer:
Density of unit cell ( rhodium) = 12.279 g/cm³
Explanation:
Given that:
The radius (r) of a rhodium atom = 135 pm
The atomic mass of rhodium = 102.90 amu
For a face-centered cubic unit cell,

where;
a = edge length.
Making "a" the subject of the formula:


a = 381.8 pm
to cm, we get:
a = 381.8 × 10⁻¹⁰ cm
However, recall that:
where;
mass of unit cell = mass of atom × numbers of atoms per unit cell
Also;


Recall also that number of atoms in a unit cell for a face-centered cubic = 4
So;

mass of unit cell = 6.83380375 × 10⁻²² g

Density of unit cell ( rhodium) = 12.279 g/cm³