Please Help! I am in desperate need Question 1
Answer the true/false questions based on the diagram above.
I have no clue where the graph is tho, my teacher did not post one.
Addition of a catalyst will change the value for ΔH
True
False

Question 2
Consider the chemical reaction equation shown.
H + H--> H-H
Is this reaction endothermic or exothermic?

a
Endothermic
b
Exothermic

Question 3

a
AB
b
BC
c
CD
d
DE

Question 4
A chemical reaction has a reaction enthalpy of −512 kJ/mol. Is it endothermic or exothermic?
a
Endothermic
b
Exothermic

Question 5
What is the energy required to evaporate two moles of liquid water given the following equations?

2H2(g) + O2(g) → 2H2O(g) ∆H = -483.6 kJ

2H2(g) + O2(g) → 2H2O(l) ∆H = -571.6 kJ

a
44.0 kJ
b
88.0 kJ
c
527.6 kJ
d
1055.2 kJ

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

your answer would be “dry”

c

explanation: