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
Among the given choices, the right statement is that computers are advantageous because it holds a larger amount of data over graphing calculators. Calculators are easier to transport and has lower power consumption. Answer here is D.
After 100years, sample is 250g
After 200 years, sample is 125g
After 300years, sample is 62.5 g
Answer: 40 grams
Explanation:
The quantity of Heat Energy (Q) required to heat a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Since Q = 93.4J
M = ?
C = 0.129 J/g.C
Φ = 40.4°C - 22.3°C = 18.1°C
Then, Q = MCΦ
Make Mass, M the subject formula
M = Q/CΦ
M = (93.4J) / (0.129 J/g.C x 18.1°C)
M = 93.4J / 2.33J/g
M = 40 g
Thus, the mass of the lead is 40 grams
Answer:
See explanation
Explanation:
The molecular geometry of an atom is connected to the number of electron pairs that surround it(whether lone pairs or bonding pairs) as well as its hybridization state. We shall now examine the N, P, or S atoms in each of the following compounds.
a)
In H3PO4, P has a tetrahedral molecular geometry and is sp3 hybridized.
b) In NH4NO3
N is sp3 hybridized in NH4^+ and sp2 hybridized in NO3^-. Also, N is tetrahedral in NH4^+ but trigonal planar in NO3^-.
c) In S2Cl2, we expect a tetrahedral geometry but as a result of the presence of two lone pairs on each sulphur atom, the molecular geometry is bent. The sulphur is sp3 hybridized.
d) In K4[O3POPO3], each phosphorus atom is in a tetrahedral molecular geometry and is sp3 hybridized.
