Consider the car driving around a circular track .Once stated,the driver maintains a speed 0f 45m/h.Compare the driver's velocit
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Answer:
a) Volume of vial= 9.626cm3
b) Mass of vial with water = 62.92 g
Explanation:
a) Mass of empty vial = 55.32 g
Mass of Vial + Hg = 185.56 g
Therefore,
Density of Hg = 13.53 g/cm3
b) Volume of water = volume of vial = 9.626 cm3
Density of water = 0.997 g/cm3
Answer:
The answer to this can be arrived at by clculating the mole fraction of atoms higher than the activation energy of 10.0 kJ by pluging in the values given into the Arrhenius equation. The answer to this is 20.22 moles of Argon have energy equal to or greater than 10.0 kJ
Explanation:
From Arrhenius equation showing the temperature dependence of reaction rates.
where
k = rate constant
A = Frequency or pre-exponential factor
Ea = energy of activation
R = The universal gas constant
T = Kelvin absolute temperature
we have
Where
f = fraction of collision with energy higher than the activation energy
Ea = activation energy = 10.0kJ = 10000J
R = universal gas constant = 8.31 J/mol.K
T = Absolute temperature in Kelvin = 400K
In the Arrhenius equation k = Ae^(-Ea/RT), the factor A is the frequency factor and the component e^(-Ea/RT) is the portion of possible collisions with high enough energy for a reaction to occur at the a specified temperature
Plugging in the values into the equation relating f to activation energy we get
or f =
= 20.22 moles of argon have an energy of 10.0 kJ or greater
We have the next % composition:
C. 49.48%
H, 5.19%.
N. 28.85%
0. 16.48%
We assume 100 g of sample
1) As we have 100 g of sample of Caffeine, we calculate the mass of each element involved here.
C. 49.48 g
H, 5.19 g
N. 28.85 g
0. 16.48 g
2) We calculate the number of moles of each element (we need the mass per mole of each element)
For C) 12.01 g/mol
49.48 g x (1 mol/12.01 g) = 4.120 moles
For H) 1.007 g/mol
5.19 g x (1 mol/1.007 g) = 5.154 moles
For O) 15.99 g/mol
16.48 g x (1 mol/15.99 g) = 1.030 moles
For N) 14.00 g/mol
28.85 g x (1 mol/14.00 g) = 2.060 moles
3) We choose the smallest number from 2) and divide the rest of them by it.
For C) 4.120 moles/1.030 moles= 4
For H) 5.154 moles/1.030 moles= 5
For O) 1.030 moles/1.030 moles= 1
For N) 2.060 moles/1.030 moles= 2
4) The numbers in 3) represents the subindex from the empirical formula of caffeine:
5) We calculate the molar mass of our empirical formula, 97.06 g/mol.
We already have the molar mass of the molecular formula, so we proceed like this:
n= the molar mass of the molecular formula/the molar mass of the empirical formula
n = 194.19 g/mol/97.06 g/mol = 2 approx.
We use "n" and we multiply our empirical formula by n = 2:
Therefore, our molecular formula: