Answer:
28.9 g
Explanation:
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.
<em>Gather all the information in one place</em> with molar masses above the formulas and masses below them.
: 159.69 28.01
Fe₂O₃ + 3CO ⟶ 2Fe + 3CO₂
Mass/g: 55.0
1. Use the molar mass of Fe₂O₃ to calculate the moles of Fe₂O₃.
2. Use the molar ratio of CO:Fe₂O₃ to calculate the moles of CO.
3.Use the molar mass of CO to calculate the mass of CO.
Answer:
A liquid-fueled rocket has two liquids (liquids are good because of the density, they need less space than a gas to be stored), such that these liquids are called the fuel and the oxidizer.
These liquids are injected into a system that leads to a combustion chamber, where the liquids are mixed (we need to mix the fuel with the oxidizer to enable the combustion of the fuel) and burned to produce thrust.
Some common examples of oxidizers are liquid oxygen, which may be combined with fuels like liquid hydrogen, liquid methane, kerosene and hydrazine.
Other oxidizers are liquid fluorine (which also can be combined with the fuels liquid hydrogen and hydrazine), nitrogen tetroxide (which can be combined whit kerosene, hydrazine and other fuels) and FLOX-70, which can only be combined with kerosene.
The "most commonly used" may depend on the country and the type of liquid propellant ( petroleum, cryogens, and hypergols)
Such that the most common oxidizer may be liquid oxygen, and the most common fuel the kerosene.
Yes. Heating up the solvent gives the molecules more kinetic energy. The more rapid motion means that the solvent molecules collide with the solute with greater frequency and the collisions occur with more force. Both factors increase the rate at which the solute dissolves.
So multiply number of moles x number of atoms/mole = 1.8066 x 10^24 atoms of H2. One mole of any gas at STP has a volume of 22.4 L. So first determine the number of moles of gas you have.
for example do 7
that 's what I think
Answer:
Distance = 200 km
Distance = 204 km
Speed = 77 km/h
Time = 21.42 h
Explanation:
Given:
A.
Speed = 100 km/h , Time = 2 h
Find:
Distance
B.
Speed = 68 km/h , Time = 3 h
Find:
Distance
C.
Distance = 154 km , Time = 2 h
Find:
Speed
D.
Distance = 1500 km speed = 70 km/h
Find:
Time
Computation:
Speed = distance / time
A.
Distance = 100 x 2
Distance = 200 km
B.
Distance = 68 x 3
Distance = 204 km
C.
Speed = 154 / 2
Speed = 77 km/h
D.
Time = 1500 / 70
Time = 21.42 h
