What is the delta H when 72.0 grams H2O condenses at 100.00C? Here are some constants that you MAY need.

specific heats heat of fusion heat of vaporization

H2O(s) = 2.1 J/g0C 6.01 kJ/mole 40.7 kJ/mole

H2O(L) = 4.18 J/g0C

H2O(g) = 1.7 J/g0C

2930 kJ

163 kJ

-163 kJ

-2930 kJ

The value of is -163 kJZ

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

......(1)

We are given:

Given mass of water = 72.0 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

Calculating the heat released for the condensation process:

......(2)

where,

= amount of heat released

n = number of moles of water = 4 moles

= specific heat of vaporization = -40.7 kJ/mol

Negative sign represents the amount of heat released.

Putting values in equation 2:

Hence, the value of is -163 kJ

hloride, what is the mass of calcium in

explanation:

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