Answer:
most likely that (2) the replicated experiment was performed incorrectly.
Why, u ask? u dare question me:
1- The initial experiment invalidness cannot be proven.
2- <em><u>t</u></em><em><u>h</u></em><em><u>e</u></em><em><u> </u></em><em><u>s</u></em><em><u>e</u></em><em><u>c</u></em><em><u>o</u></em><em><u>n</u></em><em><u>d</u></em><em><u> </u></em><em><u>a</u></em><em><u>n</u></em><em><u>s</u></em><em><u>w</u></em><em><u>e</u></em><em><u>r</u></em><em><u> </u></em><em><u>i</u></em><em><u>s</u></em><em><u> </u></em><em><u>c</u></em><em><u>o</u></em><em><u>r</u></em><em><u>r</u></em><em><u>e</u></em><em><u>c</u></em><em><u>t</u></em>
3- Different labaratories does not effect the outcome, as long as the parameter and environment of the replicated experiment is the same as when the initial experiment was conducted.
4- Already knowing the data and errors would increase the precision of the replicated experiment.
5- Change in variables should still be in the objective (or purpose) of the experiment, thus, major difference in the outcome should not happen.
happy learning!
Answer: it’s number 2 add a catalyst
Hopefully this helped :)
The energy change if 84.0 g of CaO react with excess water is 98KJ of heat is released.
calculation
heat = number of moles x delta H
delta H = - 65.2 Kj/mol
first find the number of moles of CaO reacted
moles = mass/molar mass
the molar mass of CaO = 40 + 16= 56 g/mol
mass = 84 g
moles therefore = 84 g/56 g/mol =1.5 moles
Heat is therefore = 1.5 moles x -65.2 = - 97.8 Kj = -98 Kj
since sign is negative the energy is released
Keeping the masses of the objects unchanged, if the distance between the objects is halved, then the magnitude of gravitational force between them will become. Hope this helps please mark brainliest :)
D, <span>Monotonic gases, which have no inter molecular attractions are most suited as ideal gases </span><span />
