The given elements put into an equation using their symbols are as follows:
Pb +
=
+ Ag
Since there are 2 Pb on the right side of the equation, you would change the coefficient of Pb on the left side to 2:
2Pb +
=
+ Ag
Since there are 2 Acetate on the right side of the equation, you would change the coefficient of Silver Acetate on the left side to 2:
2Pb +
=
+ Ag
Now there are 2 Silver on the left side, so you change the coefficient of Silver on the right side to 2:
2Pb +
=
+ 2Ag
That is your final equation
The coefficients are 2 + 2 = 1 + 2
Answer:
a) 88.48%
b) 0.05625 mol
Explanation:
2CH₃CH₂OH(l) → CH₃CH₂OCH₂CH₃(l) + H₂O(g) Reaction 1
CH₃CH₂OH(l) → CH₂═CH₂(g) + H₂O(g) Reaction 2
a) CH₃CH₂OH = 46.0684 g/mol
CH₃CH₂OCH₂CH₃ = 74.12 g/mol
1 mol CH₃CH₂OH ______ 46.0684 g
x ______ 50.0 g
x = 1.085 mol CH₃CH₂OH
1 mol CH₃CH₂OCH₂CH₃ ______ 74.12 g g
y ______ 35.9 g
y = 0.48 mol CH₃CH₂OCH₂CH₃
100% yield _____ 0.5425 mol CH₃CH₂OCH₂CH₃
w _____ 0.48 mol CH₃CH₂OCH₂CH₃
w = 88.48%
b) Only 0.96 mol of ethanol reacted to form diethyl ether. This means that 0.125 mol of ethanol did not react. 45% of 0.125 mol reacted to form ethylene. Therefore, 0.05625 mol of ethanol reacted by the side reaction (reaction 2). Since 1 mol of ethanol leads to 1 mol of ethylene, 0.05625 mol of ethanol produces 0.05625 mol of ethylene.
This statement is <u>TRUE.</u>
Explanation:
Due to the high heat capacity of water, it takes much more energy to have a gram of water raised by one degree compared to one gram of land. This is why during a hot day, the ground gets hotter faster than the adjacent oceans or adjacent lakes. It also takes water to lose the same amount of energy to have its temperatures drop by one degree as compared to land. During the night, therefore, this is why adjacent water bodies remain relatively warmer than land.
2Ag⁺(aq) + Mg(s)→ 2Ag(s) + Mg²⁺ (aq)
<h3>Further explanation</h3>
Given
Standard cell notation:
Mg(s) | Mg2+ (aq) || Ag+(aq)| Ag(s)
Required
a balanced redox reaction
Solution
At the cathode the reduction reaction occurs, the anode oxidation reaction occurs
In reaction:
Ag⁺ + Mg → Ag + Mg²⁺
half-reactions
- at the cathode (reduction reaction)
Ag⁺ (aq) + e⁻ ---> Ag (s) x2
2Ag⁺ (aq) + 2e⁻ ---> 2Ag (s)
- at the anode (oxidation reaction)
Mg (s) → Mg²⁺ (aq) + 2e−
a balanced cell reaction
<em>2Ag⁺(aq) + Mg(s)→ 2Ag(s) + Mg²⁺ (aq)
</em>
The acid dissociation constant is defined as Ka = [H+][A-]/[HA] where [H+], [A-] and [HA] are the concentrations of protons, conjugate base, and acid in solution respectively. Assuming this is a weak acid as the pH is quite high for a 1.35 M solution, we can assume that the change in [HA] is negligible and therefore [HA] = 1.35 M.
To calculate [H+] we can use the relationship pH = -log[H+], rearranging to give: [H+] = 10^(-pH) = 10^(-2.93) = 1.17 x 10^(-3).
Since the acid is relatively concentrated we can assume therefore that [H+] = [A-] as for each proton dissociated, a conjugate base is formed.
Therefore, we can calculate Ka as:
Ka = [H+]^2/[HA] = (1.17 x 10^-3 M)^2/1.35 = 1.01 x 10^-6 M
