Answer:
8 Silicon atom are present in unit cell.
16 oxygen atoms are present unit cell.
Explanation:
Number of atoms in unit cell = Z =?
Density of silica = tex]2.32 g/cm^3[/tex]
Edge length of cubic unit cell = a = 0.700 nm = 
Molar mass of Silica = 
Formula used :
where,
= density
Z = number of atom in unit cell
M = atomic mass
= Avogadro's number
a = edge length of unit cell
On substituting all the given values , we will get the value of 'a'.
1 silicon is 2 oxygen atoms. then 8 silicon atoms will be 16 oxygen atoms.
Answer:
The answer to your question is V2 = 825.5 ml
Explanation:
Data
Volume 1 = 750 ml
Temperature 1 = 25°C
Volume 2= ?
Temperature 2 = 55°C
Process
Use the Charles' law to solve this problem
V1/T1 = V2/T2
-Solve for V2
V2 = V1T2 / T1
-Convert temperature to °K
T1 = 25 + 273 = 298°K
T2 = 55 + 273 = 328°K
-Substitution
V2 = (750 x 328) / 298
-Simplification
V2 = 246000 / 298
-Result
V2 = 825.5 ml
Functional groups create reactive sites in molecules.
The polar part of a molecule that can hydrogen bond to water is said be hydrophilic.
Pi (π) bonds create active sites and will react with electron-deficient species.
A electronegative heteroatom like nitrogen, oxygen, or a halogen makes a carbon atom electrophilic.
(carbon will have less electronic density, which is attracted by the more electronegative heretoatoms, and it will tend to attract electron rich chemical species, and in this situation we say that the carbon atom is electrophilic).
The nonpolar part of a molecule that is not attracted to water is said to be hydrophobic.
A lone pair on a heteroatom makes it basic and nucleophilic.
(the heteroatom with the lone pair will tend to attract electron poor chemical species, and in this situation we say that the heteroatom is nucleophilic).
Answer:
The answers are explained below
Explanation:
a)
Given: concentration of salt/base = 0.031
concentration of acid = 0.050
we have
PH = PK a + log[salt]/[acid] = 1.8 + log(0.031/0.050) = 1.59
b)
we have HSO₃⁻ + OH⁻ ------> SO₃²⁻ + H₂O
Moles i............0.05...................0.01.................0.031.....................0
Moles r...........-0.01.................-0.01................0.01........................0.01
moles f...........0.04....................0....................0.041.....................0.01
c)
we will use the first equation but substituting concentration of base as 0.031 + 10ml = 0.031 + 0.010 = 0.041
Hence, we have
PH = PK a + log[salt]/[acid] = 1.8 + log(0.041/0.050) = 1.71
d)
pOH = -log (0.01/0.510) = 1.71
pH = 14 - 1.71 = 12.29
e)
Because the buffer solution (NaHSO3-Na2SO3) can regulate pH changes. when a buffer is added to water, the first change that occurs is that the water pH becomes constant. Thus, acids or bases (alkali = bases) Additional may not have any effect on the water, as this always will stabilize immediately.
