Copper (0) oside ry hydrogen
n 을
Liquid L.
T T1
pa
Corabustion tube
Ice-cold water
(1) State the observation made in the combustion tube.
(1 mark)
(in) Write down the equation leading to formation of liquid L.
(1 mark)
(iii) What property of hydrogen is being investigated.
(1 mark)
(iv) Why is potassium oxide not used to investigate this property of hydrogen gas. I mark
(v) Hydrogen gas is used in hydrogenation of oils. What do you understand by the term hydrogenation?
(vi) Give any 2 other industrial uses of hydrogen gas.
(2 marks)
9. Fractional distillation of liquid air is mainly used to obtain nitrogen and oxygen.
(i) Name one substance that is used to remove carbon (IV) oxide from the air before it is
changed into liquid
(1 mark) (ii) Describe bow
nitrogen gas is obtained from the liquid air. (Boiling points nitrogen = -196°C, Oxygen = -183°C)​

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 temperature of the gas, in k, not celsius.

5.6⋅10−^19  j

do you need the explaination?