Answer:
answer is gold
Explanation:
bcz gold is having a property of malleable so it can be drawn into thin sheets
The pH of the buffer is 6.1236.
Explanation:
The strength of any acid solution can be obtained by determining their pH. Even the buffer solution strength of the weak acid can be determined using pH. As the dissociation constant is given, we can determine the pKa value as the negative log of dissociation constant value.
![pKa=-log[H] = - log [ 5.66 * 10^{-7}]\\ \\pka = 7 - log (5.66)=7-0.753=6.247\\\\pka = 6.247](https://tex.z-dn.net/?f=pKa%3D-log%5BH%5D%20%3D%20-%20log%20%5B%205.66%20%2A%2010%5E%7B-7%7D%5D%5C%5C%20%5C%5Cpka%20%3D%207%20-%20log%20%285.66%29%3D7-0.753%3D6.247%5C%5C%5C%5Cpka%20%3D%206.247)
The pH of the buffer can be known as
![pH = pK_{a} + log[\frac{[A-]}{[HA]}}]](https://tex.z-dn.net/?f=pH%20%3D%20pK_%7Ba%7D%20%2B%20log%5B%5Cfrac%7B%5BA-%5D%7D%7B%5BHA%5D%7D%7D%5D)
The concentration of ![[A^{-}] = Moles of [A]/Total volume = 0.608/2 = 0.304 M\\](https://tex.z-dn.net/?f=%5BA%5E%7B-%7D%5D%20%3D%20Moles%20of%20%5BA%5D%2FTotal%20volume%20%3D%200.608%2F2%20%3D%200.304%20M%5C%5C)
Similarly, the concentration of [HA] = 
Then the pH of the buffer will be
pH = 6.247 + log [ 0.304/0.404]
So, the pH of the buffer is 6.1236.
Answer:
I
Explanation:
Among the halogens given in this problem, iodine has the lowest attraction for electrons.
This property is known as electronegativity.
Electronegativity is expressed as the relative tendency with which the atoms of the element attracts valence electrons in a chemical bond.
- As you go down the periodic group the electronegativity decreases.
- The most electronegative element on the periodic table is fluorine.
- Down the group, iodine is the least electronegative
- This is due to the large size of its atom.
Explanation:
Ionic equation
NaCl(aq) --> Na+(aq) + Cl-(aq)
Na2SO4(aq) --> 2Na+(aq) + SO4^2-(aq)
In NaCl solution, 1 mole of Na+ is dissociated in 1 liter of solution while in Na2SO4, 2 moles of Na+ is dissociated in 1 liter of solution.
Molecular weight of NA2SO4 = (23*2) + 32 + (16*4)
= 142 g/mol
Molecular weight of NaCl = 23 + 35.5
= 58.5 g/mol
Masses
% Mass of NA+ in Na2SO4 = mass of Na+/total mass of Na2SO4 * 100
= 46/142 * 100
= 32.4%
% Mass of NA+ in NaCl = mass of Na+/total mass of NaCl * 100
= 23/58.5 * 100
= 39.3%
Therefore, the % mass of Na+ in NaCl and Na2SO4 are different so it cannot be used.
Hi,
Gases are most soluble in water under high pressure and low pressure.
