In 1905 Albert Einstein had proposed a solution to the problem of observations made on the behaviour of light having characteristics of both wave and particle theory. From work of Plank on emission of light from hot bodies, Einstein suggested that light is composed of tiny particles called <span>photons, </span>and each photon has energy.
Light theory branches in to the physics of <span>quantum mechanics, </span>which was conceptualised in the twentieth century. Quantum mechanics deals with behaviour of nature on the atomic scale or smaller.
As a result of quantum mechanics, this gave the proof of the dual nature of light and therefore not a contradiction.
Hydrogen is usually –1. This is INCORRECT. The oxidation number for H is +1.
Oxygen is usually –2. This is CORRECT.
A pure group 1 element is +1. This is INCORRECT. It does not follow. This will depend on the other elements and the overall charge.
A monatomic ion is 0. This is INCORRECT. Diatomic ion is 0.
Answer:
1.60.
Explanation:
- The no. of millimoles of HCl = MV = (0.15 M)(20.0 mL) = 3.0 mmol.
- The no. of millimoles of KOH = MV = (0.10 M)(20.0 mL) = 2.0 mmol.
<em>Since the no. of millimoles of HCl is larger than that of KOH. The solution is acidic.</em>
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∴ M of remaining HCl [H⁺] remaining = (NV)HCl - (NV)KOH/V total = (3.0 mmol) - (2.0 mmol) / (40.0 mL) = 0.025 M.
∵ pH = - log[H⁺]
<em>∴ pH = - log[H⁺] </em>= - log(0.025) = <em>1.602 ≅ 1.60.</em>
Answer:
The answer to your question is below
Explanation:
Factors that affect the rate of a chemical reaction
- Temperature If the temperature increases the rate of reaction increases.
- Concentration The reaction will move where there less concentration it could be to the reactants of products.
- Particle size The lower the particle size the higher the rate of reaction.
- Catalyst Catalyzers accelerate the rate of reaction
- Pressure The reaction will move where there are fewer molecules.
Answer:
Explanation:
<u>1) Rate law, at a given temperature:</u>
- Since all the data are obtained at the same temperature, the equilibrium constant is the same.
- Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:
r = K [A]ᵃ [B]ᵇ
<u>2) Use the data from the table</u>
- Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:
r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s
r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s
Divide r₂ by r₁: [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0
- Use the first and second set of data to find the exponent a:
r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s
r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s
Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]
2ᵃ = 2 ⇒ a = 1
<u>3) Write the rate law</u>
This means, that the rate is independent of reactant B and is of first order respect reactant A.
<u>4) Use any set of data to find K</u>
With the first set of data
- r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹
Result: the rate constant is K = 0.167 s⁻¹
