So here's how you find the answer:
Given: (rate constants)
K₁ = 4.0 x 10⁻⁴ M⁻¹s⁻¹.
T₁ = 25.0 C = 293 K.
k₂ = 2.6 x10⁻³ M⁻¹s⁻¹.
T₂ = 42.4 C = 315.4 K.
R = 8,314 J/K·mol.
Use the equation:
ln(k₁/k₂) = Ea/R (1/T₂ - 1/T₁).
Transpose:
Ea = R·T₁·T₂ / (T₁ - T₂) · ln(k₁/k₂)
Substitute within the given transposed equation:
<span>Ea = 8,314 J/K·mol · 293 K · 315.4 K ÷ (293 K - 315.4 K) · ln (4.0 x 10⁻⁴ M⁻¹s⁻¹/ 2.6 x 10⁻³ M⁻¹s⁻¹).
</span>
Continuing the solution we get:
<span>Ea = 768315 J·K/mol ÷ (-22,4 K) * (-1.87)
</span>
The value of EA is:
<span>Ea = 64140.58 J/mol ÷ 1000 J/kJ = 64.140 kJ/mol.</span>
Answer:
The other signal that would indicate the presence of a C= C bond appears close to 3100 
.
Explanation:
Bands that appear above 3000 are often unsaturation diagnoses suggest. The band at 3000-
3100 is characteristics for C-H stretching frequencies and normally is overlaps with the ones for alkanes because it is a band of weak intensity.
94.20 g/3.16722 mL = 29.74 g/mL
The ratio of mass to volume is equal to the substance's density. Thus, 29.74 g/mL is the density of whatever substance it may be. Density does not change for incompressible matter like solid and some liquids. Although, it may be temperature dependent.
<u>Answer:</u> The initial amount of Uranium-232 present is 11.3 grams.
<u>Explanation:</u>
All the radioactive reactions follows first order kinetics.
The equation used to calculate half life for first order kinetics:
We are given:
Putting values in above equation, we get:
Rate law expression for first order kinetics is given by the equation:
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = 
t = time taken for decay process = 206.7 yrs
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= initial amount of the reactant = ?
[A] = amount left after decay process = 1.40 g
Putting values in above equation, we get:
![0.0101yr^{-1}=\frac{2.303}{206.7yrs}\log\frac{[A_o]}{1.40}](https://tex.z-dn.net/?f=0.0101yr%5E%7B-1%7D%3D%5Cfrac%7B2.303%7D%7B206.7yrs%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B1.40%7D)
![[A_o]=11.3g](https://tex.z-dn.net/?f=%5BA_o%5D%3D11.3g)
Hence, the initial amount of Uranium-232 present is 11.3 grams.
The balanced chemical reaction is written as:
<span>Cu +2AgNO3 → Cu(NO3)2 + 2Ag
</span>
We are given the amount of silver nitrate to be used for the reaction. This value will be the starting point of our calculations. It is as follows:
4.00 g AgNO3 ( 1 mol / 169.87 g ) ( 1 mol Cu / 2 mol AgNO3 ) ( 63.456 g / 1 mol ) = 0.747 g Cu
