Answer:
Scientific theories are testable and make falsifiable predictions. They describe the causes of a particular natural phenomenon and are used to explain and predict aspects of the physical universe or specific areas of inquiry (for example, electricity, chemistry, and astronomy).
A good theory in the theoretical sense is (1) consistent with empirical observations; is (2) precise, (3) parsimonious, (4) explanatorily broad, and (5) falsifiable; and (6) promotes scientific progress (among others; Table 1.1).
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
Kinetic Energy which relies on an objects mass and velocity and Potential Energy which relies on the height of the object
Answer:
The answer is (B.) Chlorine
Explanation:
Its because i took the test
Answer:
a) [H3O+] = 1.00 E-10 M ⇒ [OH-] = 1.0 E-4 M
b) [H3O+] = 1.00 E-4 M ⇒ [OH-] = 1.0 E-10 M
c) [H3O+] = 9.90 E-6 M ⇒ [OH-] = 1.0 E-9 M
Explanation:
- 14 = pH + pOH
- pH = - Log [H3O+]
a) [H3O+] = 1.00 E-10 M
⇒ pH = - Log(1.00 E-10) = 10
⇒ pOH = 14 - 10 = 4
⇒ 4 = - Log[OH-]
⇒ [OH-] = 1.0 E-4 M
b) [H3O+] = 1.00 E-4 M
⇒ pH = 4
⇒ pOH = 10
⇒ [OH-] = 1.0 E-10 M
c) [H3O+] = 9.90 E-6 M
⇒ pH = 5
⇒ pOH = 9
⇒ [OH-] = 1.0 E-9 M
