Answer:
(a) ΔU = 7.2x10²
(b) W = -5.1x10²
(c) q = 5.2x10²
Explanation:
From the definition of power (p), we have:
(1)
<em>where, p: is power (J/s = W (watt)) W: is work = ΔU (J) and t: is time (s) </em>
(a) We can calculate the energy (ΔU) using equation (1):
(b) The work is related to pressure and volume by:

<em>where p: pressure and ΔV: change in volume = V final - V initial </em>
(c) By the definition of Energy, we can calculate q:
<em>where Δq: is the heat transfer </em>
I hope it helps you!
<h2>Acetic Acid + Sodium ethoxide ⇄ Butyric Acid + Sodium Hydroxide</h2>
Explanation:
An ionic equation for the reaction of acetic acid with sodium ethoxide is as follows -
Acetic Acid and Sodium ethanolate (sodium ethoxide)
Butyric Acid and Sodium hydroxide
Hence,
Acetic Acid + Sodium ethoxide ⇄ Butyric Acid + Sodium Hydroxide
⇄ 
- Weak acids and bases have low energy than strong acids and bases.
- The chemical equilibria shift the reaction side with the species having lower energy.
- Given reaction is an acid-base reaction in which the equilibrium favors the starting material that means it will go to the side of the weakest acid that is acetic acid is weaker than butyric acid.
Answer:
Life
Explanation:
With out him we wouldn't live
Answer:
The answer to your question is Final volume = 58.37 ml
Explanation:
Data
density = 8.96 g/cm³
mass = 75 g
volume of water = 50 ml
Process
1.- Calculate the volume of copper
Density = mass / volume
Solve for volume
Volume = mass / density
Substitution
Volume = 75/8.96
Simplification
Volume = 8.37cm³ or 8.37 cm³
2.- Calculate the new volume of water in the graduated cylinder
Final volume = 50 + 8.37
Final volume = 58.37 ml
Answer:
0.062mol
Explanation:
Using ideal gas law as follows;
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821Latm/molK)
T = temperature (K)
Based on the information provided;
P = 152 Kpa = 152/101 = 1.50atm
V = 0.97L
n = ?
T = 12°C = 12 + 273 = 285K
Using PV = nRT
n = PV/RT
n = (1.5 × 0.97) ÷ (0.0821 × 285)
n = 1.455 ÷ 23.39
n = 0.062mol