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
Answer:
m = 180 g
Explanation:
Given data:
Energy absorbed = 108 J
Mas of gold = ?
Initial temperature = 25°C
Final temperature = 29.7 °C
Specific heat capacity of gold = 0.128 J/g.°C
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT =29.7 °C - 25°C
ΔT = 4.7 °C
108 J = m ×0.128 J/g.°C ×4.7 °C
108 J = m ×0.60 J/g
m = 108 J/0.60 J/g
m = 180 g
Answer:
The one with the greatest mass would be the one that has the most things in the nucleus, protons and nutrons
Explanation:
Calculate the mass of the solute <span>in the solution :
Molar mass KCl = </span><span>74.55 g/mol
m = Molarity * molar mass * volume
m = 0.9 * 74.55 * 3.5
m = 234.8325 g
</span><span>To prepare 0.9 M KCl solution, weigh 234.8325 g of salt in an analytical balance, dissolve in a beaker, shortly after transfer with the help of a funnel of transfer to a volumetric flask of 100 cm</span>³<span> and complete with water up to the mark, then cover the balloon and finally shake the solution to mix
hope this helps!</span>
It is important to take note of th temperature in determining the density of a substance because this will set as a basis and will likely be a variable in the experiment because this will also contribute on the effects of the experiment and a basis of how the experiment has turned to be that way.
