Answer:
i think its true but nevermind its true✌
<span><span>When water vapor condenses, 2260 joules/gram heat energy will be released into the atmosphere.
To add, </span>heat energy<span> <span>(or </span>thermal energy<span> or simply </span>heat) is defined as a form of energy<span> which transfers among particles in a substance (or system) by means of kinetic </span>energy<span> of those particles. In other words, under kinetic theory, the </span>heat<span> is transferred by particles bouncing into each other.</span></span></span>
Answer:
Part a: The rate of the equation for 1st order reaction is given as ![Rate=k[H_2O_2]](https://tex.z-dn.net/?f=Rate%3Dk%5BH_2O_2%5D)
Part b: The integrated Rate Law is given as ![[H_2O_2]=[H_2O_2]_0 e^{-kt}](https://tex.z-dn.net/?f=%5BH_2O_2%5D%3D%5BH_2O_2%5D_0%20e%5E%7B-kt%7D)
Part c: The value of rate constant is 
Part d: Concentration after 4000 s is 0.043 M.
Explanation:
By plotting the relation between the natural log of concentration of
, the graph forms a straight line as indicated in the figure attached. This indicates that the reaction is of 1st order.
Part a
Rate Law
The rate of the equation for 1st order reaction is given as
![Rate=k[H_2O_2]](https://tex.z-dn.net/?f=Rate%3Dk%5BH_2O_2%5D)
Part b
Integrated Rate Law
The integrated Rate Law is given as
![[H_2O_2]=[H_2O_2]_0 e^{-kt}](https://tex.z-dn.net/?f=%5BH_2O_2%5D%3D%5BH_2O_2%5D_0%20e%5E%7B-kt%7D)
Part c
Value of the Rate Constant
Value of the rate constant is given by using the relation between 1st two observations i.e.
t1=0, M1=1.00
t2=120 s , M2=0.91
So k is calculated as

The value of rate constant is 
Part d
Concentration after 4000 s is given as

Concentration after 4000 s is 0.043 M.
Given that the drum is replaced each time before sampling, the selections are independent events. Then, the probability of the combined event is the product of each probability: (50/5000) * (50/5000) * (50/5000) = (1/100)^3 = 0.000001
Butane undergoes combustion to produce carbon dioxide and water, expressed in the following reaction <span> C4H10 (g) + O2 (g) → 4CO2 (g) + 5H2O (g) . 1 mole of butane is to 4 moles of CO2. In this case, when 22.8 grams of butane or 0.393 moles of it, 69.18 grams of carbon dioxide is produced</span>