The question mentions a change in temperature from 25 to 50 °C. With that, the aim of the question is to determine the change in volume based on that change in temperature. Therefore this question is based on Gay- Lussac's Gas Law which notes that an increase in temperature, causes an increase in pressure since the two are directly proportional (once volume remains constant). Thus Gay-Lussac's Equation can be used to solve for the answer.
Boyle's Equation:
=
Since the initial temperature (T₁) is 25 C, the final temperature is 50 C (T₂) and the initial pressure (P₁) is 103 kPa, then we can substitute these into the equation to find the final pressure (P₂).
=
∴ by substituting the known values, ⇒ (103 kPa) ÷ (25 °C) = (P₂) ÷ (50 °C)
⇒ P₂ = (4.12 kPa · °C) (50 °C)
=
206 kPa
Thus the pressure of the gas since the temperature was raised from 25 °C to 50 °C is
206 kPa
This question is missing the part that actually asks the question. The questions that are asked are as follows:
(a) How much of a 1.00 mg sample of americium remains after 4 day? Express your answer using 2 significant figures.
(b) How much of a 1.00 mg sample of iodine remains after 4 days? Express your answer using 3 significant figures.
We can use the equation for a first order rate law to find the amount of material remaining after 4 days:
[A] = [A]₀e^(-kt)
[A]₀ = initial amount
k = rate constant
t = time
[A] = amount of material at time, t.
(a) For americium we begin with 1.00 mg of sample and must convert time to units of years, as our rate constant, k, is in units of yr⁻¹.
4 days x 1 year/365 days = 0.0110
A = (1.00)e^((-1.6x10^-3)(0.0110))
A = 1.0 mg
The decay of americium is so slow that no noticeable change occurs over 4 days.
(b) We can simply plug in the information of iodine-125 and solve for A:
A = (1.00)e^(-0.011 x 4)
A = 0.957 mg
Iodine-125 decays at a much faster rate than americium and after 4 days there will be a significant loss of mass.
- E(Bonds broken) = 1371 kJ/mol reaction
- E(Bonds formed) = 1852 kJ/mol reaction
- ΔH = -481 kJ/mol.
- The reaction is exothermic.
<h3>Explanation</h3>
2 H-H + O=O → 2 H-O-H
There are two moles of H-H bonds and one mole of O=O bonds in one mole of reactants. All of them will break in the reaction. That will absorb
- E(Bonds broken) = 2 × 436 + 499 = 1371 kJ/mol reaction.
- ΔH(Breaking bonds) = +1371 kJ/mol
Each mole of the reaction will form two moles of water molecules. Each mole of H₂O molecules have two moles O-H bonds. Two moles of the molecule will have four moles of O-H bonds. Forming all those bond will release
- E(Bonds formed) = 2 × 2 × 463 = 1852 kJ/mol reaction.
- ΔH(Forming bonds) = - 1852 kJ/mol
Heat of the reaction:
is negative. As a result, the reaction is exothermic.
Explanation:
high energy to Low energy
=the electron gains energy (K.E)
The Reaction is spontaneous when temperature is 430 K. Hence, Option (C) is correct.
<h3>
</h3><h3>
What is Spontaneous reaction ?</h3>
Reactions are favorable when they result in a decrease in enthalpy and an increase in entropy of the system.
When both of these conditions are met, the reaction occurs naturally.
Spontaneous reaction is a reaction that favors the formation of products at the conditions under which the reaction is occurring.
According to Gibb's equation:
ΔG = ΔH - TΔS
ΔG = Gibbs free energy
ΔH = enthalpy change = +62.4 kJ/mol
ΔS = entropy change = +0.145 kJ/molK
T = temperature in Kelvin
- ΔG = +ve, reaction is non spontaneous
- ΔG = -ve, reaction is spontaneous
- ΔG = 0, reaction is in equilibrium
ΔH - TΔS = 0 for reaction to be spontaneous
T = ΔH / ΔS
Here,
T = 500K
Thus the Reaction is spontaneous when temperature is 500 K.
Learn more about Gibbs free energy here ;
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