1. for dry air at 1. atm pressure, the densities at –50°c, 0°c, and 69°c are 1.5826 g dm–3 , 1.2929 g dm–3, and 1.0322 g dm–3, respectively. a) assume a sample of mass 1000 g, and calculate the volume at each temperature. b) from these data, and assuming that air obeys charles’s law, determine a value for th

Answer :

(a) The value of volume of air at temperature are respectively.

Explanation :

(a) The formula used for density is:

or,

Now we have to calculate the volume at each temperature.

The mass of sample = 1000 g

At temperature :

Density =

At temperature :

Density =

At temperature :

Density =

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