<span>C6H12 = 6x12 + 6x1 = 78.
The equation indicates that 2x78 = 156g benzene will produce 6542kJ.
Using proportions you can then calculate that
x/6542kJ = 7.9g / 156g
x = 331.3kJ = 331300J.
heat = mass x ΔT x 4.18J/g°
ΔT = 331300J / (5691g x 4.18J/g°) = 13.9°
final temp = 21 + 14° = 35°C</span>
Answer:
Explanation:
Given that:
The argon atoms are excited into an excited state before emitting the 488.0 nm laser.
the energy of the first ionization energy of argon is 1520 kJ mol-1.
SInce 1 eV = 96.49 kJ/mol
Therefore, the energy of the first ionization energy of argon in eV is = ( 1520/ 96.49) eV
= 15.75 eV
To find where the energy level of the excited state lies below the vacuum energy level, let's first determine, the energy liberated by using planck expression.
Converting Joules (J) to eV ; we get,
E = 2.53 eV
The energy levels of the first exited state = -13.223 eV
8.9 g/cm3 since density equals mass/volume or m/v you just divide 89 by 10 to get 8.9
Answer:
1.21 times
Explanation:
The energy of a wave is proportional to the square of the amplitude of the wave.
Mathematically:
where
E is the energy of the wave
A is its amplitude
In this problem, the amplitude of the wave increases by a factor of 1.1; it means that the new amplitude can be written as
Therefore, this means that the energy of the wave increases by a factor:
Therefore, the energy of the wave increases by a factor 1.21.
You need to add the last substance to the products side(the right sode of the arrow). You have hydrogen and oxygen - water.
You get: BrO3 + N2H4 -> Br2 + N2 + H2O
# of Br: 1x1 = 1 # of Br: 2x1 = 2
O: 3x1 = 3 O: 1x1 = 1
N: 2x1 = 2 B N: 2x1 = 2
H: 4x1 = 4. H: 2x1 = 2
Br:
Multiply the reactant (left) side by 2 to balance.
O:
You've just multiplied the reactant oxygen by 2 so now the reactant side equals 6. Multiply the product (right) side by six as well.
H:
The product side is now equal to 12. Multiply the reactant side by 3 to balance.
N:
Now you have to balance N because the reactant side has been risen. So multiply the product side by three as well.
You end up with the complete and balanced equation:
2BrO3 + 3N2H4 -> Br2 + 3N2 + 6H2O