A 500-cm3 sample of 1.0 M NaOH(aq) is added to 500 cm3 of 1.0 M HCl(aq) in a Styrofoam cup, and the solution is quickly stirred. The rise in temperature (ΔT1) is measured. The experiment is repeated using 100 cm3 of each solution, and the rise in temperature (ΔT2) is measured. What conclusion can you draw about ΔT1 and ΔT2?HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq); ΔH° = –55.8 kJ

explanation:

0.07 g/cm³

1.37 g/cm³

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.