Answer:
I think you can predict it by the number of protons in it.
Answer:
wwe are asked to prive the atomic pavking factor APF for BCC is 0.68
solution
APF is given by equation (3.3) in the form
APF =nVs/Vc
whenn is the number of atoms per unit cell and for BCC n=2.0, Vs is the volume of the atom and it is given by 43π3³ and Vc is the volume of the unit cell and for BCC Vc=a³ where a is the side of the cube. From equation 3.4
we could get a by
a=4R/√3
where R is the radius of the atom. Now let us olug the expression inyo equation (1) to get APF
APF = nVs/Vc
=2.0(4/3πR³) / (4R/√3)³
=2.0(4/3πR³) / (4/√3)³ R³
=0.68
Hello!
Data:
Molar Mass of H2CO3 (carbonic acid)
H = 2*1 = 2 amu
C = 1*12 = 12 amu
O = 3*16 = 48 amu
------------------------
Molar Mass of H2CO3 = 2 + 12 + 48 = 62 g/mol
Now, since the Molarity and ionization constant has been supplied, we will find the degree of ionization, let us see:
M (molarity) = 0.01 M (Mol/L) →
Use: Ka (ionization constant) =
Now, we will calculate the amount of Hydronium [H3O+] in carbonic acid (H2CO3), multiply the acid molarity by the degree of ionization, we will have:
And finally, we will use the data found and put in the logarithmic equation of the PH, thus:
Data:
apply the data to formula
Note:. The pH <7, then we have an acidic solution (weak acid).
Now, let's find pOH by the following formula:
I Hope this helps, greetings ... DexteR! =)
Given parameters:
Weight of hydrated sample = 20g
Temperature = 250°C
Weight after cooling = 16.5g
Unknown:
Weight of water lost from the sample = ?
Solution:
The weight of water lost from the sample;
Weight of water lost = Weight of hydrated sample - Weight of dry sample
Weight of water lost = 20g - 16.5g = 3.5g
% of water in the sample =
Input parameters solve;
=
=17.5%