Answer:
Exercise 1
a) 140 m
b) 100 m
c) 180 m
d) 140 m
Exercise 2
a) 20 yards
b) 30 yards
c) 20 yards
d) 55 yards
Exercise 3
a) 11 Kilometers
b) 7 Kilometers
Explanation:
Exercise 1
Distance from B to C = 140 m
Distance from to D = 100 m
Total distance = 100+40+40 = 180 m
Total displacement i.e. distance between A and D is 140 m
Exercise 2
a) Distance from B to C = 35 -15 = 20 yards
b) Distance from C to D = 5 + 35 = 30 yards
c) Distance from B to D = 5 + 15 = 20 yards
d) Displacement = 5 + 50 = 55 yards
Exercise 3
a) The total distance travelled = 5 + 2 + 4 = 11 Kilometers
b) Displacement = 5-2 + 4 = 7 Kilometers
Answer:
d the amount of matter in a closed system is the same at the start of a reaction as at the end of the reaction
Answer:
. Name a soluble compound that could be added to precipitate all of the iron ions from the solution.
Sodium Hydroxide.
Beta radiation - negative charge and the mass of an electron
Gamma radiation- no mass and no charge
Alpha radiation- He 4+/2
Answer:
The percentage yield is 78.2g
Explanation:
Given, mass of propane = 42.8 g , sufficient O2 percent yield = 61.0 % yield.
Reaction - C3H8(g)+5O2(g)------> 3CO2(g)+4H2O(g)
First we need to calculate the moles of propane
Moles of propane = 
g.mol-1
= 0.971 moles
So, moles of CO2 from the moles of propane
1 mole of C3H8(g) = 3 moles of CO2(g)
So, 0.971 moles of C3H8(g) = ?
= 2.913 moles of CO2
So theoretical yield = 2.913 moles 
44.0 g/mol
= 128.2 g
So, the actual mass of CO2 = percent yield theoretical yield / 100 %
= 61.0 % 128.2 g / 100 %
= 78.2 g
the mass of CO2 that can be produced if the reaction of 42.8 g of propane and sufficient oxygen has a 61.0 % yield is 78.2 g
