The density of a substance can simply be calculated by
dividing the mass by the volume:
density = mass / volume
Therefore calculating for the density since mass and volume
are given:
density = 46.0 g / 34.6 mL
density = 1.33 g / mL
Answer:
The concentration is 50,8 % w/v and radio strengths = 1,96.
Explanation:
Phenobarbital sodium is a medication that could treat insomnia, for example.
2,0 M of Phenobarbital sodium means 2 moles in 1L.
The concentration units in this case are %w/v that means 1g in 100 mL and ratio strengths that means 1g in <em>r</em> mL. Thus, 2 moles must be converted in grams with molar weight -254 g/mole- and liters to mililiters -1 L are 1000mL-. So:
2 moles ×
= 508 g of Phenobarbital sodium.
1 L ×
= 1000 mL of solution
Thus, % w/v is:
× 100 = 50,8 % w/v
And radio strengths:
= 1,96. Thus, you have 1 g in 1,96 mL
I hope it helps!
Answer : The number of bonding electrons and the number of non-bonding electrons are (4, 18).
Explanation :
The number of bonding electrons and non-bonding electrons in the structure of
is determined by the Lewis-dot structure.
Lewis-dot structure : It tell us about the number of valence electrons of an atom within a molecule and it is also shows the bonding between the atoms of a molecule and the lone-pair of electrons.
In the given structure, 'Xe' is the central atom and 'F' is the terminal atom.
Xenon has 8 valence electrons and fluorine has 7 valence electrons.
Total number of valence electrons in
= 8 + 2(7) = 22 electrons
From the Lewis-dot structure, we conclude that
The number of electrons used in bonding = 4
The number of electrons used in non-bonding (lone-pairs) = 22 - 4 = 18
Therefore, the number of bonding electrons and the number of non-bonding electrons are (4, 18).
The Lewis-dot structure of
is shown below.
Answer:


Explanation:
first write the equilibrium equaion ,
⇄ 
assuming degree of dissociation
=1/10;
and initial concentraion of
=c;
At equlibrium ;
concentration of
![[C_3H_5O_3^{-} ]= c\alpha](https://tex.z-dn.net/?f=%5BC_3H_5O_3%5E%7B-%7D%20%20%5D%3D%20c%5Calpha)
![[H^{+}] = c\alpha](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%20c%5Calpha)

is very small so
can be neglected
and equation is;

= 
![P_H =- log[H^{+} ]](https://tex.z-dn.net/?f=P_H%20%3D-%20log%5BH%5E%7B%2B%7D%20%5D)





composiion ;
![c=\frac{1}{\alpha} \times [H^{+}]](https://tex.z-dn.net/?f=c%3D%5Cfrac%7B1%7D%7B%5Calpha%7D%20%5Ctimes%20%5BH%5E%7B%2B%7D%5D)
![[H^{+}] =antilog(-P_H)](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3Dantilog%28-P_H%29)
![[H^{+} ] =0.0014](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%20%5D%20%3D0.0014)

