Answer:
3.09kg
Explanation:
First, let us write a balanced equation for the reaction. This is illustrated below:
2C8H18 + 25O2 —> 16CO2 + 18H2O
Molar Mass of C8H18 = (12x8) + (18x1) = 96 + 18 = 114g/mol
Mass of C8H18 from the balanced equation = 2 x 114 = 228g
Converting 228g of C8H18 to kg, we obtained:
228/1000 = 0.228kg
Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol
Mass of CO2 from the balanced equation = 16 x 44 = 704g
Converting 704g of CO2 to kg, we obtained:
704/1000 = 0.704kg
From the equation,
0.228kg of C8H18 produced 0.704kg of CO2.
Therefore, 1kg of C8H18 will produce = 0.704/0.228 = 3.09kg of CO2
We have get the mass of gaseous water after evaporation in a closed container.
The mass of water vapor after evaporation is 5 grams.
In closed container, there is no exchange in mass from system to surrounding, only heat may exchange. The number of moles of water vapour remains unchanged as 5 gram water is heated in closed container.
Due to heating, liquid water gets evaporated and intermolecular distance between water molecules increases in gaseous state than liquid state and intermolecuar force of attraction decreases.
Randomness of molecules increases in gaseous state than liquid state.
The number of moles of a gas if it occupy a volume of 4dm at 10 atm and - 200 c is 6.67 moles
calculation
by use of ideal gas equation
that is PV =nRT
where P(pressure) = 10 atm
V( volume) = 4dm^3 = 4L
n(number of moles) =?
R(gas constant) = 0.0821 L.atm/mol.K
T = -200 +273 = 73 K
make n the formula of the subject n = PV/RT
n is therefore = (10 atm x 4l)/(0.0821 L.atm/mol.K x 73 K) = 6.67 moles
Matter :)))))))) or mass depends
You'd take about ten or eleven drops per day