A. Density of Liquids
A.1 Density of Water
1. Volume (mL) 20.0ml
2. Mass of beaker 36.09g
3. Mass of beaker+liquid 56.22g
4. Mass of liquid
5. Density

A.2 Density of an Unknown Liquid aim
1. Type of liquid sample ethanol
2. Volume of liquid 20.0 ml
3. Mass of beaker 52.89g
4. Mass of beaker +liquid
5. Mass of liquid
6. Density

B. Specific Gravity
1. Specific gravity of water
2. Specific gravity of liquid sample
3. Specific gravity of water using a hydrometer 1.02
4. Specific gravity of liquid using a hydrometer 0.791

How do the calculated specific gravity of water and the liquid sample (1, 2) compare to the hydrometer readings (3, 4) for water and the liquid sample?

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

the cold front replaces the warmer air at the ground level changing the wind to be breezy and denser.there is also an immediate drop in the temperature.