Answer:
The given molecules are SO2 and BrF5.
Explanation:
Consider the molecule SO2:
The central atom is S.
The number of domains on S in this molecule is three.
Domain geometry is trigonal planar.
But there is a lone pair on the central atom.
So, according to VSEPR theory,
the molecular geometry becomes bent or V-shape.
Hybridization on the central atom is
.
Consider the molecule BrF5:
The central atom is Br.
The number of domains on the central atom is six.
Domain geometry is octahedral.
But the central atom has a lone pair of electrons.
So, the molecular geometry becomes square pyramidal.
The hybridization of the central atom is .
The shapes of SO2 and BrF5 are shown below:
Let's begin with the basic values that will be used in the solution.
The formula of propane is C3H8. It is an alkane, a hydrocarbon with the general formula of CnH2n+2. Notice that hydrocarbons have only Carbon and Hydrogen atoms. Its molar mass (M) is 44 g.
Molar Mass Calculation is done as like that
C=12 g/mol, H=1 g/mol. 1 mole propane has 3 moles Carbon atoms and 8 mole Hydrogen atoms. M(C3H8)= 3*12+ 8*1= 44 g
Combustion reaction of hydrocarbons gives carbon dioxide and water by releasing energy. That energy is called as enthalpy of combustion (ΔHc°).
ΔHc° of propane equals -2202.0 kj/mol. Burning of 1 mole C3H8 releases 2202 kj energy. Minus sign only indicates that the energy is given out ( an exothermic reaction ).
Let's write the combustion reaction.
C3H8 + O2 ---> CO2 + H20 (unbalanced) ΔHc° = -2202 kj/mol
Now, we calculate mole of 20 kg propane. Convert kilogram into gram since we use molar mass is defined in grams.
mole=mass/molar mass ; n=m/M ; n= 20000 g /44 (g/mol)=454 mole
1 mole propane releases 2202 kj energy.
454 mole propane release 2202 kj *454= 1000909 kj
The answer is 1000909 kj.
Answer:
hope this helped :( got this from my friend a long time ago but still not sure if it's really correct this
Explanation:
The given data is as follows.
Volume of lake = =
Concentration of lake = 5.6 mg/l
Total amount of pollutant present in lake =
= mg
= kg
Flow rate of river is 50
Volume of water in 1 day =
= liter
Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are or
Flow rate of sewage =
Volume of sewage water in 1 day = liter
Concentration of sewage = 300 mg/L
Total amount of pollutants = or
Therefore, total concentration of lake after 1 day =
= 6.8078 mg/l
= 0.2 per day
= 6.8078
Hence, =
=
= 1.234 mg/l
Hence, the remaining concentration = (6.8078 - 1.234) mg/l
= 5.6 mg/l
Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.
Answer:
Well I know that when the liquid changes from clear to dark blue and small bubbles form and rise to the surface that is a reaction
Explanation: