Every mole of CH4 used, three moles of H2 are produced, so 2 moles of CH4, would be 6 moles of H2 produced
Answer: Flammability is a material's ability to burn in the presence of oxygen.
Explanation: Chemical properties can be observed only when the substance changes into one or more different substances through chemical reactions or transformations. One of the chemical properties is flammability.
Flammability is a material's ability to burn in the presence of oxygen.
Remember, oxygen doesn't burn. Precisely flammable substances obtain substances that burn. Oxygen remains an oxidizing agent, which means it supports the combustion process. Oxygen causes other objects to catch fire at low temperatures and burns hotter and faster. But oxygen itself does not burn. Consequently, if you at present deliver fuel and fire, adding oxygen will provide the fire.
Carbon dioxide is the result of combustion. An example can be seen in firewood in a fireplace. One of the chemical properties of carbon-based wood is having the ability to burn. Chemically the wood turns into carbon dioxide when it burns and leaves a residue of ash. Furthermore, this ash residue cannot be turned back into the wood. Chemical changes result in new substances.
Consider an example of a combustion reaction to methane gas:
Our balanced equation for methane combustion implies that every one CH₄ molecule reacts with two O₂ molecules. The product of combustion is one carbon dioxide molecule and two steam or water vapor molecules.
Answer:
C) C4H6 - Right answer
Explanation:
Let's combine the Ideal Gases Law with density to get the molecular formula for the unknown gas.
Density = mass / volume
1.06 g /L means that 1.06 grams of compound occupy 1 liter of volume.
P . V = n . R . T
Pressure in Torr must be converted to atm
760 Torr are 1 atm
371 Torr are __ (371 .1)/760 = 0.488 atm
0.488 atm . 1L = 1.06g/MM . 0.082 . 304K
(0.488 atm . 1L) / 0.082 . 304K = 1.06g/ MM
Mass / Molar mass = Moles → That's why the 1.06 g / MM
0.0195 mol = 1.06g / MM
1.06g/0.0195 mol = MM → 54.3 g/m
Now, let's use the composition
100 g of compound have 88.8 g of C
54.3 g of compound have ___ (54.3 . 88.8) /100 = 48 g of C
100 g of compound have 11.2 g of H
54.3 g of compound have __ (54.3 . 11.2)/100 = 6 g of H
48 g of C are included un 4 atoms
6 g of H are included in 6 atoms
Answer:
B. Spring balance - a device used for measuring the weight or force of gravity acting on an object.
Explanation:
A Force is any interaction that changes the motion or position of an obkpjectbthatbit is interacting with. Whenever there is an interaction between two objects, there is a force exerted by each of the objects on one themselves.
Forces are generally divided into contact forces and non-contact over field forces.
In contact forces, the two objects physically in contact with each other. Examples of contact forces are push or pull forces, frictional forces, tensional forces, spring forces, etc.
Non-contact forces are forces in which the two objects interacting do no need to be physically in contact with one another. Examples include, gravitational forces, magnetic forces, electrical forces, etc.
Instruments used in measuring forces are known as force gauges.
From the instruments listed above:
A. A ruler is an instrument used in measuring length
B. Spring balance is a device used for measuring the weight or force of gravity acting on an object.
C. A thermometer is an instrument used in measuring temperature
D. A windbvane is an instrument used in measuring wind direction.
Answer:
molar composition for liquid
xb= 0.24
xt=0.76
molar composition for vapor
yb=0.51
yt=0.49
Explanation:
For an ideal solution we can use the Raoult law.
Raoult law: in an ideal liquid solution, the vapor pressure for every component in the solution (partial pressure) is equal to the vapor pressure of every pure component multiple by its molar fraction.
For toluene and benzene would be:
Where:
is partial pressure for benzene in the liquid
is benzene molar fraction in the liquid
vapor pressure for pure benzene.
The total pressure in the solution is:
And
Working on the equation for total pressure we have:
Since 
We know P and both vapor pressures so we can clear from the equation.
So
To get the mole fraction for the vapor we know that in the equilibrium:
So
Something that we can see in these compositions is that the liquid is richer in the less volatile compound (toluene) and the vapor in the more volatile compound (benzene). If we take away this vapor from the solution, the solution is going to reach a new state of equilibrium, where more vapor will be produced. This vapor will have a higher molar fraction of the more volatile compound. If we do this a lot of times, we can get a vapor that is almost pure in the more volatile compound. This is principle used in the fractional distillation.
