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:
10−8 M.
Explanation:
In this problem we are given pH and asked to solve for the hydrogen ion concentration. Using the equation, pH = − log [H+] , we can solve for [H+] as,
− pH = log [H+] ,
[H+] = 10−pH,
by exponentiating both sides with base 10 to "undo" the common logarithm. The hydrogen ion concentration of blood with pH 7.4 is,
[H+] = 10−7.4 ≈ 0.0000040 = 4.0 × In this problem we are given pH and asked to solve for the hydrogen ion concentration. Using the equation, pH = − log [H+] , we can solve for [H+] as,
− pH = log [H+] ,
[H+] = 10−pH,
by exponentiating both sides with base 10 to "undo" the common logarithm. The hydrogen ion concentration of blood with pH 7.4 is,
[H+] = 10−7.4 ≈ 0.0000040 = 4.0 × 10−8 M.
Answer:
Explanation:
You would have to add up the atomic masses of all the compounds in the compound, making sure you include how many molecules of each are in the compound
For example, in CuSOA we have 1 molecule of Cu and S, as 4 molecules of O
The atomic masses are as follows:
Cu = 63.55 u
S = 32.065 u
O = 15.99 units
This is how we would add it up:
(Atomic mass of Cu) + (Atomic mass of S) + 4(Atomic Mass of O)
(63.55) + (32.065) + 4(15.99)
(63.55) + (32.065) + 63.96
= 159.575 u
