Answer is periodic because only happens in certain times
Answer:
Explanation:
C + O2 → CO2
Mole of C = 24 g/(12 g/mole)
Mole of C = 2 mole
Mole of molecular O2 = 74 g/(32 g/mole)
Mole of molecular O2 = 2.3125 mole
Since mole of C < mole of O2, then C being the limiting reagent.
From the reaction, it shows that mole ratio between C and O2 = 1 : 1.
So, 2 moles of C will stoichiometrically react with 2 moles of O2 to generate 2 moles of CO2.
Avogadro's law states that :"equal volumes of all gases, at the same temperature and pressure, have the same number of molecules i.e. 6.02 x 10^23 molecules/mole.
Therefore, 2 moles of CO2 contain 2 moles x 6.02 x 10^23 molecules/mole = 1.204 x 10^24 molecules of CO2 is formed.
Answer:
1) 0,081 ft/s
2) 0,746 lb/s
Explanation:
The relation between flow and velocity of a fluid is given by:
Q=Av
where:
- Q, flow [ft3/s]
- A, cross section of the pipe [ft2]
- v, velocity of the fluid [ft/s]
1)
To convert our data to appropiate units, we use the following convertion factors:
1 ft=12 inches
1 ft3=7,48 gallons
1 minute=60 seconds
So,
As the pipe has a circular section, we use A=πd^2/4:
Finally:
Q=vA......................v=Q/A
2)
The following formula is used to calculate the specific gravity of a material:
SG = ρ / ρW
where:
- ρ = density of the material [lb/ft3]
- ρW = density of water [lb/ft3] = 62.4 lbs/ft3
then:
ρ = SG*ρW = 1,49* 62,4 lb/ft3 = 93 lb/ft3
To calculate the mass flow, we just use the density of the chloroform in lb/ft3 to relate mass and volume:
In a particular experiment, the per cent yield is 79.0%. This means that in this experiment, a 7.90-g sample of fluorine yields is 7g of SF6.
<h3>How is Sulphur hexafluoride formed?</h3>
Sulfur Hexafluoride is a disparity agent formed of an inorganic fluorinated inert gas comprised of six fluoride atoms bound to one sulfur atom, with possible diagnostic activity upon imaging.
Thus, a sample of fluorine yields 7g of SF6.
To learn more about Sulfur Hexafluoride click here;
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The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
