Student z completes his/her calibration step for part 3 and finds an average massa to be 95.237 g ± 0.005 g. however, in his/her calculations, s/he mistakenly uses 85.237 g ± 0.005 g for massa. s/he uses the correct values for massb and masscyl. will the calculated mass of displaced water be too high, too low, or unchanged? and why?

- Calculated mass of displaced water would be too low.

Explanations:- Mass of water displaced is directly proportional to the mass of the object dropped into water.

Since the average mass of A used for calculations is low than the actual mass, the calculated mass of displaced water would also be low.

answer is: nuclear fusion produces elements that are heavier than helium.

nuclei combine to form a heavier nucleus, releasing energy.  

for example, fusion of hydrogen atoms at the sun.  

this fusion nuclear reaction is in the inner core of the sun and hydrogen (atomic number 1) is conveted into helium (atomic number 2).  

the inner core is the central region of the sun and all solar energy is produced in the core by nuclear fusion.  

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.