C. quadruples the rate
<h3>Further explanation</h3>
Given
The rate law :
R=k[A]²
Required
The rate
Solution
There are several factors that influence reaction kinetics :
- 1. Concentration
- 2. Surface area
- 3. Temperature
- 4. Catalyst
- 5. Pressure
- 6. Stirring
The rate is proportional to the concentration.
If the concentration increased, the reaction rate will increase
The reaction is second-order overall(The exponent is 2)
The concentration of A is doubled, the reaction rate will increase :
r = k[A]² ⇒ r= k[2A]²⇒r=4k[A]²
<em>The reaction rate will quadruple.</em>
Answer:
2.33g of iron (iii) chloride
50.0 mL of 5.00 M of sodium phosphate
FeCl3 + Na3PO4 > Fe(PO4) + 3NaCl
mol = conc × vol = 0.5 × 50/1000 = 0.025 mol Na3PO4
from the equation:
1 mol of Na3PO4 reacts with 1 mol FeCl3 = 3 mol of NaCl
0.025 mol = x
x = 0.0025 × 3 = 0.075 mol NaCl
mass = 0.075 g × 59 g/mol = 4.425 g NaCl
i guessed all of this so i dont know i it is correct
I think because its the only one to be liquid at normal temperatures.
(a) One form of the Clausius-Clapeyron equation is
ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂); where in this case:
Solving for ΔHv:
- ΔHv = R * ln(P₂/P₁) / (1/T₁ - 1/T₂)
- ΔHv = 8.31 J/molK * ln(5.3/1.3) / (1/358.96 - 1/392.46)
(b) <em>Normal boiling point means</em> that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, <u>solving for T₂</u>:
- ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂)
- 1/T₂ = 1/T₁ - [ ln(P₂/P₁) / (ΔHv/R) ]
- 1/T₂ = 1/358.96 K - [ ln(101.325/1.3) / (49111.12/8.31) ]
(c)<em> The enthalpy of vaporization</em> was calculated in part (a), and it does not vary depending on temperature, meaning <u>that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK</u>.
