Answer:
second law of thermodynamics.
Explanation:
The second law of thermodynamics deals with interconversion of energy from one form to another. Although energy can be converted from one form to another, this conversion is never 100% efficient because energy is lost in certain ways such as through heat. In a combustion engine, it is not possible to recover the energy from the gasoline 100% since energy must be lost along the way via such means as heat losses. Hence I will be skeptical about such an advert.
It take more energy to break the bonds of the reactants and less energy is given off when the product bonds are formed.
<h3>What is Energy?</h3>
Energy is defined as the ability to do work. Work is done in the breaking or formation of bonds.
The standard Enthalpy (ΔH) of water which was formed in the given reaction is negative.
ΔH= Δproduct - Δreactant
This means that the energy to break the bonds of the reactants is more.
Read more about Enthalpy here brainly.com/question/14291557
Answer:
See below
Explanation:
<u> Name </u> <u>Formula </u> <u> Major species </u> <u> </u>
Zinc iodide ZnI₂ H₂O(ℓ), I⁻(aq), Zn²⁺(aq),
Nitrogen(I) oxide N₂O H₂O(ℓ), N₂O(aq)
Sodium nitrite NaNO₂ H₂O(ℓ), Na⁺(aq), NO₂⁻(aq)
Glucose C₆H₁₂O₆ H₂O(ℓ), C₆H₁₂O₆(aq)
Nickel(II) iodide NiI₂ H₂O(ℓ), I⁻(aq), Ni²⁺(aq)
- Glucose and nitrogen(I) oxide are covalent compounds. They do not dissociate in solution.
- The compounds containing metals are ionic. They produce ions in solution.
- ZnI₂ and NiI₂ produce twice as many iodide ions as metal ions.
Answer:
The net ionic equation is
C6H5COOH+ CN-= C6H5COO- + HCN
Explanation:
From the ionic equation
C6H5COOH + Na+ + CN- = C6H5COO- + Na+ + HCN
Only sodium is the spectator ion, so it cancels out, since C6H5COOH and HCN do not ionize completely they are left undissociated
Given:
M = 0.0150 mol/L HF solution
T = 26°C = 299.15 K
π = 0.449 atm
Required:
percent ionization
Solution:
First, we get the van't Hoff factor using this equation:
π = i MRT
0.449 atm = i (0.0150 mol/L) (0.08206 L atm / mol K) (299.15 K)
i = 1.219367
Next, calculate the concentration of the ions and the acid.
We let x = [H+] = [F-]
[HF] = 0.0150 - x
Adding all the concentration and equating to iM
x +x + 0.0150 - x = <span>1.219367 (0.0150)
x = 3.2905 x 10^-3
percent dissociation = (x/M) (100) = (3.2905 x 10-3/0.0150) (100) = 21.94%
Also,
percent dissociation = (i -1) (100) = (</span><span>1.219367 * 1) (100) = 21.94%</span>