To calculate the <span>δ h, we must balance first the reaction:
NO + 0.5O2 -----> NO2
Then we write all the reactions,
2O3 -----> 3O2 </span><span>δ h = -426 kj eq. (1)
O2 -----> 2O </span><span>δ h = 490 kj eq. (2)
NO + O3 -----> NO2 + O2 </span><span>δ h = -200 kj eq. (3)
We divide eq. (1) by 2, we get
</span>O3 -----> 1.5O2 δ h = -213 kj eq. (4)
Then, we subtract eq. (3) by eq. (4)
NO + O3 -----> NO2 + O2 δ h = -200 kj
- (O3 -----> 1.5 O2 δ h = -213 kj)
NO -----> NO2 - 0.5O2 δ h = 13 kj eq. (5)
eq. (2) divided by -2. (Note: Dividing or multiplying by negative number reverses the reaction)
O -----> 0.5O2 <span>δ h = -245 kj eq. (6)
</span>
Add eq. (6) to eq. (5), we get
NO -----> NO2 - 0.5O2 δ h = 13 kj
+ O -----> 0.5O2 δ h = -245 kj
NO + O ----> NO2 δ h = -232 kj
<em>ANSWER:</em> <em>NO + O ----> NO2 δ h = -232 kj</em>
Answer:
λ = 6.5604 x 1016 nm
Explanation:
Given Data:
The energy of the red line in Hydrogen Spectra = 3.03 x 10-19
Formula to calculate Wave length
E= hv
Where E is Energy
h is Planks Constant = 6.626 x 10–34 J s
v is frequency
In turn
v= c/ λ
where c is speed of light = 3.00 x 108 m s–1
λ is wavelength = to find
Solution:
Formula to be Used:
E= hv………………………… (1)
Putting the value v in equation 1
E= h c/ λ…………………… (2)
Put the value in equation 2
3.03 x 10-19 J = (6.626 x 10–34 J s) x (3.00 x 108 m s–1) / λ ……………………….(3)
By rearranging equation 3
λ = (6.626 x 10–34 J s) x (3.00 x 108 m s–1) /3.03 x 10-19 J
λ = 6.5604 x 107 m
The answer is in “m”
So we have to convert it into nm
So for this to convert “m” to “nm” multiply the answer with 109
λ = 6.5604 x 107 x 109
λ = 6.5604 x 1016 nm
Answer:
Explanation:
If the reaction is really exothermic (and it is) then the water would spatter all over the place. It would boil off if the container could hold it. It would also react according to the following reaction.
You are talking about a reaction like
2K + 2HOH = 2KOH + H2
Answer:
a) 1.866 × 10 ⁻¹⁹ J b) 3.685 × 10⁻¹⁹ J
Explanation:
the constants involved are
h ( Planck constant) = 6.626 × 10⁻³⁴ m² kg/s
Me of electron = 9.109 × 10 ⁻³¹ kg
speed of light = 3.0 × 10 ⁸ m/s
a) the Ek ( kinetic energy of the dislodged electron) = 0.5 mu²
Ek = 0.5 × 9.109 × 10⁻³¹ × ( 6.40 × 10⁵ )² = 1.866 × 10 ⁻¹⁹ J
b) Φ ( minimum energy needed to dislodge the electron ) can be calculated by this formula
hv = Φ + Ek
where Ek = 1.866 × 10 ⁻¹⁹ J
v ( threshold frequency ) = c / λ where c is the speed of light and λ is the wavelength of light = 358.1 nm = 3.581 × 10⁻⁷ m
v = ( 3.0 × 10 ⁸ m/s ) / (3.581 × 10⁻⁷ m ) = 8.378 × 10¹⁴ s⁻¹
hv = 6.626 × 10⁻³⁴ m² kg/s × 8.378 × 10¹⁴ s⁻¹ = 5.551 × 10⁻¹⁹ J
5.551 × 10⁻¹⁹ J = 1.866 × 10 ⁻¹⁹ J + Φ
Φ = 5.551 × 10⁻¹⁹ J - 1.866 × 10 ⁻¹⁹ J = 3.685 × 10⁻¹⁹ J
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