When a radioactive decay occurs, protons/neutrons, alpha/beta particles and gamma rays are all released depending on the type of decay.
Answer : The standard enthalpy of formation of ethylene is, 52.4 kJ
Explanation :
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The formation reaction of
will be,

The intermediate balanced chemical reaction will be,
(1)

(2)

(3)

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equation, we get :
(1)

(2)

(3)

The expression for enthalpy of formation of
will be,



Therefore, the standard enthalpy of formation of ethylene is, 52.4 kJ
Variables we know:
t = 8 seconds
Vi = 0 m/s
g = -9.81
Δy = ?
Vf = ?
Equation we will be using to solve for Vf: Vf = Vi + gt
Steps to solve:
Vf = (0) + (-9.81)(8)
Vf = -78.48 m/s
Hope this helps!! :)
Explanation:
here's the answer to your question
Answer : The labs were unable to reproduce the pharmaceutical company’s data.
Explanation : Any scientific claim must have reproducible experimental data. In this case, when the pharmaceutical company has the claim of reducing the cancer growth cells by 35% then by using the same manufacturing procedure for the drug and lab should be able to get this result. But they failed to match up with the results which clearly indicates that the labs were not able to produce the same results and hence they concluded that the pharmaceutical company's claims were invalid.