How do you find half life like i know the formula with 1/2 but how do you know how many times you do it
2 answers:
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Answer:
Y = 92.5 %
Explanation:
Hello there!
In this case, since the reaction between lead (II) nitrate and potassium bromide is:
Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:
Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:
And the resulting percent yield:
Regards!
Answer : The enthalpy change for the process is 52.5 kJ/mole.
Explanation :
Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the solution
where,
q = heat released by the reaction
= heat absorbed by the calorimeter
= heat absorbed by the solution
= specific heat of calorimeter =
= specific heat of water =
= mass of water or solution =
= change in temperature =
Now put all the given values in the above formula, we get:
Now we have to calculate the enthalpy change for the process.
where,
= enthalpy change = ?
q = heat released = 2626.61 J
n = number of moles of copper sulfate used =
Therefore, the enthalpy change for the process is 52.5 kJ/mole.