Suppose you were calibrating a 100.0 ml volumetric flask using distilled water. the flask temperature was at 20°c, and you assumed that the distilled water was as well. however, you later discover that the actual water temperature was 11°c instead. how is the mass of the 100.0 ml of distilled water you measured at 11°c different from the mass of 100.0 ml of distilled water at 20°c?

Density is sensitive to temperature for gases and liquids, although not much for liquids. We use the data in the picture. Using linear interpolation, we determine the densities at 11°C and 20°C.

@20°C: Density = 0.99823 g/cm³ or g/mL
@11°C:
(10 - 11)/(10 - 20) = (0.99973 - Density)/(0.99973 - 0.99823)
Solving for density:
Density = 0.99958 g/cm³ or g/mL

Mass @ 20°C = 100 mL *  0.99823 g/mL = 99.823 g
Mass @ 11°C = 100 mL *  0.99958 g/mL = 99.958 g
Difference of Masses = |99.823 g - 99.958 g| = 0.135 g

Hairy thick gross thing? idk

Hbr dissociates to form h+ and br-ca(oh)2 dissociates to form ca2+ and 2oh-the h+ and oh- react to form waterassuming the reactants are dissolved in water, the net reaction is h+ + oh- —> h2oif you’re asked to show all components of the reaction, the complete reaction is 2hbr + ca(oh)2 —> cabr2 + 2h2o