Answer:
salt bridge balances the charge when electrons move from one half cell to another half cell.
Explanation:
Explanation: A salt bridge balances the charge when electrons move from one half cell to another half cell. During this process the salt bridge uses its electrolyte solution which further helps in balancing charges in both the half cells. ... Therefore, for each electrochemical cell a new salt bridge is used.
Answer:
For part (a): pHsol=2.22
Explanation:
I will show you how to solve part (a), so that you can use this example to solve part (b) on your own.
So, you're dealing with formic acid, HCOOH, a weak acid that does not dissociate completely in aqueous solution. This means that an equilibrium will be established between the unionized and ionized forms of the acid.
You can use an ICE table and the initial concentration ofthe acid to determine the concentrations of the conjugate base and of the hydronium ions tha are produced when the acid ionizes
HCOOH(aq]+H2O(l]⇌ HCOO−(aq] + H3O+(aq]
I 0.20 0 0
C (−x) (+x) (+x)
E (0.20−x) x x
You need to use the acid's pKa to determine its acid dissociation constant, Ka, which is equal to
Answer:
2.25g of NaF are needed to prepare the buffer of pH = 3.2
Explanation:
The mixture of a weak acid (HF) with its conjugate base (NaF), produce a buffer. To find the pH of a buffer we must use H-H equation:
pH = pKa + log [A-] / [HA]
<em>Where pH is the pH of the buffer that you want = 3.2, pKa is the pKa of HF = 3.17, and [] could be taken as the moles of A-, the conjugate base (NaF) and the weak acid, HA, (HF). </em>
The moles of HF are:
500mL = 0.500L * (0.100mol/L) = 0.0500 moles HF
Replacing:
3.2 = 3.17 + log [A-] / [0.0500moles]
0.03 = log [A-] / [0.0500moles]
1.017152 = [A-] / [0.0500moles]
[A-] = 0.0500mol * 1.017152
[A-] = 0.0536 moles NaF
The mass could be obtained using the molar mass of NaF (41.99g/mol):
0.0536 moles NaF * (41.99g/mol) =
<h3>2.25g of NaF are needed to prepare the buffer of pH = 3.2</h3>
Elements of Group 1 and group 2 in the periodic
table contain elements so reactive that they are never found in the free state
<u>Explanation</u>:
The metals in group 1 of periodic table consisting of 'alkali metals' which include lithium, potassium, sodium, rubidium, Francium and caesium. They are highly reactive because they have low ionisation energy and larger radius. The group 2 metals consist of 'alkaline earth metals' which include calcium, strontium, barium, beryllium, radium and magnesium. These alkaline earth metal have +2 oxidation number, hence are highly reactive.
These both group metals are mostly reactive and so are never found in a free state. When they are exposed to air they would immediately react with oxygen. Hence, are stored in oils to avoid oxidation.
Answer:
3937 inches
Explanation:
1m = 39.37 inches.
However, we want to know how long a 100m soccer field is, so we have to times the value for 1m by 100.
Therefore, an 100m soccer field = 100 x 39.37 = 3937 inches
