a) pto₂ + 2h₂ ⟶ pt + 2h₂o
b) 0.021 g
c) 0.010 mol; 0.18 g
a) balanced equation
we know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
m_r: 2.016 195.08 18.02
pto₂ + 2h₂ ⟶ pt + 2h₂o
b) mass of h₂
(i) calculate the moles of pt
n = 1.0 g pt × (1 mol pt /195.08 g pt)
= 5.13 × 10⁻³ mol pt
(ii) calculate the moles of h₂
the molar ratio is (2 mol h₂/1 mol pt)
n = 5.13 × 10⁻³ mol pt × (2 mol h₂/1 mol pt)
= 0.0103 mol h₂
(iii) calculate the mass of h₂
m = 0.0103 mol h₂× (2.016 g h₂/1 mol h₂)
m = 0.021 g h₂
you need 0.021 g h₂ to produce 1.00 g pt.
c) moles of water
the molar ratio is (2 mol h₂o/1 mol pt)
moles of h₂o = 5.13 × 10⁻³ mol pt × (2 mol h₂o/1 mol pt)
= 0.010 mol h₂o
mass of water = 0.010 mol h₂o × (18.02 g h₂o/1 mol h₂o)
= 0.18 g h₂o
for an object with positive acceleration, the graph is a straight line but it has a slope. the graph is made between speed (on x-axis) and time (on y-axis). the diagram is attached.
note:if the slope of the graph is high i.e. steep graph, it means the acceleration is very high. if the slope of the graph is low, it means the acceleration is less.