d. Fe(s) and Al(s)
<h3>Further explanation</h3>
In the redox reaction, it is also known
Reducing agents are substances that experience oxidation
Oxidizing agents are substances that experience reduction
The metal activity series is expressed in voltaic series
<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au </em>
The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent
The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent
So that the metal located on the left can push the metal on the right in the redox reaction
The electrodes which are easier to reduce than hydrogen (H), have E cells = +
The electrodes which are easier to oxidize than hydrogen have a sign E cell = -
So the above metals or metal ions will reduce Pb²⁺ (aq) will be located to the left of the Pb in the voltaic series or which have a more negative E cell value (greater reduction power)
The metal : d. Fe(s) and Al(s)
Answer:
Option B. At pH extremes, the amino acid molecules mostly carry a net charge, thus increasing their solubility in polar solvent.
C. At very low or very high pH, the amino acid molecules have increased charge, thus form more salt bonds with water solvent molecules.
Explanation:
Answer:
C6H12O6 —> 2C2H5OH + 2CO2
Explanation:
The equation for the reaction is given below:
C6H12O6 —> C2H5OH + CO2
We can balance the equation above as follow:
There are 12 atoms of H on the left side and 6 atoms of the right side. It can be balance by putting 2 in front of C2H5OH as shown below:
C6H12O6 —> 2C2H5OH + CO2
There are 6 atoms of C on the left side and 5 atoms on the right side. It can be balance by putting 2 in front of CO2 as shown below:
C6H12O6 —> 2C2H5OH + 2CO2
Now the equation is balanced.
For the desired pH of 5.76, 0.365 mol of acetate and 0.035 mol of acid are to be added
let the concentration of acetate be x
then the concentration of acid will be (0.8 - x)
pKa of acetate buffer = 4.76
pH = pKa + log([acetate]/[acid])
⇒4.76 = 4.76 + log(x/(0.8-x))
⇒log(x/(0.8-x)) = 0
⇒x/(0.8-x) = 1
⇒x = 0.4
Therefore
[acetate] = x = 0.4
[acid] = 0.8-x =0.4 M
number of mol = concentration *(volume in mL)
number of mol of acetate = 0.4*0.5
= 0.20 mol
number of mol acid = 0.4*0.5
= 0.20 mol
when desired pH = 5.76
pH = pKa + log([acetate]/[acid])
⇒5.76 = 4.76 + log(x/(0.8-x))
⇒log(x/(0.8-x)) = 1
⇒x/(0.8-x) = 10
⇒x = 8 - 10x
⇒x = 8/11
⇒x= 0.73
[acetate] = x= 0.73
[acid] = 0.8-x = 0.07 M
number of mol = concentration * (volume in mL)
number of mol acetate to be added = 0.73*0.5 = 0.365 mol
number of mol acid to be added = 0.07*0.5 = 0.035 mol
Problem based on acetic acid required to maintain a certain pH
brainly.com/question/9240031
#SPJ4
To determine whether the amount of H2 in the lab is dangerous, we first need to know how much hydrogen gas is present in the room in units of percent by volume. For this particular problem, we cannot exactly determine since we do not know the total volume of the room. Hope this answers the question.
